UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


6, 


< 


AN   INTRODUCTION   TO 

THE   MODERN 
THEORY   OF  EQUATIONS 


BOOKS  BY  FLORIAN  CAJORI 


HISTORY  OF  MATHEMATICS 

Revised  and  Enlarged  Edition 

HISTORY  or  ELEMENTARY  MATHEMATICS 
Revised  and  Enlarged  Edition 

HISTORY  OF  PHYSICS 

INTRODUCTION  TO  THE  MODERN 
THEORY  OF  EQUATIONS 


AN  INTRODUCTION  TO 

THE   MODERN 

THEORY  OF  EQUATIONS 


BY 

FLORIAN   CAJORI,   PH.D. 

PROFESSOR  OF  MATHEMATICS  AT  COLORADO  COLLEGE 


Nefo  got* 
THE   MACMILLAN   COMPANY 

LONDON:  MACMILLAN  &  CO.,  LTD. 
1921 

All  rights  reserved 


COPYRIGHT,  1904, 
BY  THE  MACMILLAN  COMPANY. 


Set  up  and  eiectrotyped.     Published  October,  15 


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Norwood,  Mass.,  U.S.A. 


Sciences 

;  »  Library 

Gv  A 
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C 

PREFACE 

THE  main  difference  between  this  text  and  others  on  the 
same  subject,  published  in  the  English  language,  consists  in 
the  selection  of  the  material.  In  proceeding  from  the  ele- 
mentary to  the  more  advanced  properties  of  equations,  the 
subject  of  invariants  and  covariants  is  here  omitted,  to  make 
room  for  a  discussion  of  the  elements  of  substitutions  and 
substitution-groups,  of  domains  of  rationality,  and  of  their 
application  to  equations.  Thereby  the  reader  acquires  some 
familiarity  with  the  fundamental  results  011  the  theory  of 
equations,  reached  by  Gauss,  Abel,  Galois,  and  Kronecker. 

The  Galois  theory  of  equations  is  usually  found  by  the 
beginner  to  be  quite  difficult  of  comprehension.  In  the  pres- 
ent text  the  effort  is  made  to  render  the  subject  more  concrete 
by  the  insertion  of  numerous  exercises.  If,  in  the  work  of 
the  class  room,  this  text  be  found  to  possess  any  superiority, 
it  will  be  due  largely  to  these  exercises.  Most  of  them  are 
my  own ;  some  are  taken  from  the  treatises  named  below. 

In  the  mode  of  presentation  I  can  claim  no  originality. 
The  following  texts  have  been  used  in  the  preparation  of  this 
book: 

BACHMANN,  P.     Kreistheiluny.     Leipzig,  1872. 

BORNSIDE,  W.     Theory  of  Groups.     Cambridge,  1897. 

BURNSIDE,  W.  S.,  and  PANTON,  A.  W.     Theory  of  Equations,  Vol.  I, 
1899  ;  Vol.  II,  1901. 

DICKSOX,  L.  E.     Theory  of  Algebraic  Equations.     New  York,  1903. 

EASTON,  B.  S.     The  Constructive  Development  of  Group-Theory.     Phila- 
delphia, 1902. 

Encyklopadie  der  Mathematischen  Wissenschaften. 

v 


m 
vi 

GALOIS,  D'EVARISTE.     (Euvres  mathematiques,  avec  une  introduction  par 

M.  EMILE  PICARD.     Paris,  1897. 

KLEIN,  F.     Vorlesungen  iiber  das  Ikosaeder.     Leipzig,  1884. 
MATTHIESSEN,  L.     Grundziige  der  Antiken  u.  Modernen  Algebra.     Leip- 
zig, 1878. 
NETTO,  E.      Theory  of  Substitutions,  translated  by  F.  N.  COLE,  Ann 

Arbor,  1892.  . 
NETTO,  E.     Vorlesungen  iiber  Algebra.     Leipzig,  Vol.  I,  1896 ;  Vol.  II, 

1900. 
PETERSEN,  J.      Theorie  der  Algebraischen  Gleichungen.     Kopenhagen, 

1878. 

PIERPONT,  J.     Galois'1  Theory  on  Algebraic  Equations.     Salem,  1900. 
SALMON,  G.     Modern  Higher  Algebra.     Dublin,  1876. 
SERRET,  J.  A.     Handbuch  der  Hoheren  Algebra.     Deutsche  Uebers.    v. 

G.  WERTHEIM.     Leipzig,  1878. 

TODHUNTER,  I.     Theory  of  Equations.     London,  1880. 
VOGT,  H.     Resolution  Algebrique  des  Equations.     Paris,  1895. 
WEBER,    H.      Lehrbuch   der  Algebra.      Braunschweig,    Vol.    I,    1898 ; 

Vol.  II,  1896. 
WEBER,   H.      Encyklopadie   der   Elementaren  Algebra   und  Analysis. 

Leipzig,  1903. 

Of  these  books,  some  have  been  used  more  than  others.  In 
the  elementary  parts  I  have  been  influenced  by  the  excellent 
treatment  found  in  the  first  volume  of  Burnside  and  Panton. 
In  the  presentation  of  the  Galois  theory  I  have  followed  the 
first  volume  of  Weber's  admirable  Lehrbuch  der  Algebra.  Next 
to  these,  special  mention  of  indebtedness  is  due  to  Bachmaun, 
Netto,  Serret,  and  Pierpont. 

I  desire  also  to  express  my  thanks  to  Miss  Edith  P.  Hub- 
bard,  of  the  Cutler  Academy,  Miss  Adelaide  Denis,  of  the  Col- 
'orado  Springs  High  School,  and  Mr.  R.  E.  Powers,  of  Denver, 
for  valuable  suggestions  and  assistance  in  the  reading  of  the 
proofs,  and  to  Mr.  W.  N.  Birchby,  who  has  furnished  solutions 
to  a  large  number  of  problems. 

FLORIAN  CAJORI. 

COLORADO  COLLEGE, 
January,  1904. 


TABLE   OF   CONTENTS 

CHAPTER   I 

PAGE 

SOME  ELEMENTARY  PROPERTIES  OF  EQUATIONS,  §§  1-26          .         .         1  v> 

CHAPTER   II 
ELEMENTARY  TRANSFORMATIONS  OF  EQUATIONS,  §§  27-36        .         ,       31  r 

CHAPTER   III 
LOCATION  OF  THE  ROOTS  OF  AN  EQUATION,  §§  37-51       .        .  43  ^ 

CHAPTER   IV 
APPROXIMATION  TO  THE  ROOTS  OF  NUMERICAL  EQUATIONS,  §§  52-58      60  c 

CHAPTER   V 
THE  ALGEBRAIC  SOLUTION  OF  THE  CUBIC  AND  QUARTIC,  §§  59-62      68    k'; 

CHAPTER   VI 

SOLUTION  OF  BINOMIAL  EQUATIONS  AND   RECIPROCAL  EQUATIONS, 

§§63-67 74  <T" 

CHAPTER   VII 

SYMMETRIC  FUNCTIONS  OF  THE  ROOTS,  §§  68-71      ....      84 

CHAPTER   VIII 
ELIMINATION,  §§  72-77 92 


Vlll  TABLE   OF   CONTENTS 

CHAPTER   IX 

PAGB 
THE     HOMOGRAPHIC    AND     THE     TSCHIRXHAUSEN    TRANSFORMATIONS, 

§§  78-80 99 

CHAPTER   X 

ON  SUBSTITUTIONS,  §§  81-93 104 

>  / 

CHAPTER  XI 
SUBSTITUTION-GROUPS,  §§  94-113       .......    112 

CHAPTER   XH 
RESOLVENTS  OF  LAGRANGB,  §§  114-119 129 

CHAPTER   XIII 

THE   GALOIS  THEORY   OF   ALGEBRAIC   NUMBERS.      REDUCTIBILITY, 

§§  120-139 134 

O  o 

CHAPTER   XIV 
NORMAL  DOMAINS,  §§  140-159 150 

CHAPTER   XV 
REDUCTION  OF  THE  GALOIS  RESOLVENT  BY  ADJUNCTION,  §§  160-166     174 

CHAPTER   XVI 

THE  SOLUTION  OF  EQUATIONS  VIEWED  FROM  THE   STANDPOINT  OF 

THE  GALOIS  THEORY,  §§  167-169 184 

CHAPTER  XVII 
CYCLIC  EQUATIONS,  §§  170-183 187 


TABLE   OF   CONTENTS  IX 


CHAPTER   XVIII 

PAGE 

ABELIAN  EQUATIONS,  §§  184-189 210 


CHAPTER   XIX 

THE  ALGEBRAIC  SOLUTION  OF  EQUATIONS,  §§  190-201     .         .         ,     219 


(r 


THEOEY   OF   EQUATIONS 

CHAPTER   I 

SOME  ELEMENTARY  PROPERTIES  OF  EQUATIONS 

1.  Functions.  In  the  study  of  the  theory  of  equations  we 
shall  employ  a  class  of  functions  called  algebraic.  An  algebraic 
function  is  one  which  involves  only  the  operations  of  addition, 
subtraction,  multiplication,  division,  involution,  and  evolution 
in  expressions  with  constant  exponents.  Thus,  x2  +  ax  +  b, 

V2  x2  +  1,  — - —  are  examples  of  algebraic  functions ;    while 
x  +  5 

sin  y,  ex,  log  (1  +  x),  tan"1  z  are  examples  of  functions  which 
are  not  algebraic,  but  transcendental. 

A  rational  function  of  a  quantity  is  one  which  involves  only 
the  operations  of  addition,  subtraction,  multiplication,  and 
division  upon  that  quantity.  If  root-extraction  with  respect 
to  any  operand  containing  that  quantity  is  involved,  then  the 
function  is  irrational.  An, .integral,  function  of  a  quantity  is 
one  in  which  the  quantity  never  appears  in  the  denominator  of 
a  fraction.  Thus,  ay2  +  by  +  c  is  a  rational,  y*  +  ?/-+!  is  an 
irrational  function  of  y ;  f  y?  +  \  x  is  an  integral  function  of 

x,  while  -  is  not  an  integral  function.     The  expression  /(#), 

X 

defined  thus, 

f(x)  =  a^n  +  ajcc"-1  +  a^x"-2  -i 1-  <*„_!»  +  an,  I 

is  a  rational  integral  algebraic  function  of  x  of  the  nth  degree,  n 
being  assumed  to  be  a  positive  integer.     The  coefficients  «0>  ai> 

B  1 


2  THEORY   OF   EQUATIONS 

a.,,  •  ••,  a,,  are  numbers  independent  of  x.     A  variety  of  furthel 
assumptions  relating  to  these  coefficients  may  be  made. 

Thus,  we  may  assume  that  they  are  variables,  varying  inde- 
pendently of  each  other.  It  will  be  seen  that,  in  this  case,  the 
roots  of  the  equation  f(x)  =  0  are  quantities  independent  of 
each  other.  We  may  also  assume  that  the  variable  coefficients 
are  rational  functions  of  one  or  more  other  variables.  Thus,  in 
tot?  +  t~x  -\-  (t2  -f  £),  the  coefficients  are  functions  of  the  variable  t. 

Or,  we  may  assume  the  coefficients  to  be  constants  —  either 
particular  algebraic  numbers  or  letters  which  stand  for  such 
numbers. 

-The  nature  of  the  assumptions  relating  to  the  coefficients 
will  be  stated  definitely  as  we  proceed.  In  some  theorems  the 
coefficients  are  confined  to  real,  rational,  integral  numbers ;  in 
others,  the  coefficients  may  be  fractions  or  complex  numbers; 
in  the  development  of  the  Galois  Theory  of  Equations,  radical 
expressions  will  be  admitted.  But  in  no  case  are  the  coeffi- 
cients supposed  to  be  transcendental  numbers,  such  as  TT  or 
e  =  2.718  .... 

Whenever,  in  the  next  ten  chapters,  the  coefficients  are  rep- 
resented by  letters,  they  may  be  regarded  either  as  independent 
variables  or  as  constants.  Not  until  we  enter  upon  the  Galois 
theory  is  it  essential  to  discriminate  between  the  two. 

2.  The  equation  obtained  by  putting  the  polynomial  I  in  §  1 
equal  to  zero  is  called  an  algebraic  equation  of  the  nth  derjr<><>. 
We  designate  it   briefly  by  /(«•)  =  0.     A  value   of   x   which 
reduces  this  equation__to_jin  identity_is_called  a  root. 

When  all  the  coefficients  are  independent  variables,  the  equa- 
tion is  the  so-called  general  equation  of  the  nth  degree.  Viewed 
from  the  standpoint  of  the  Galois  theory,  it  will  be  seen,  §  111, 
that  the  so-called  general  equation  is  not  the  true  general  case, 
but  really  only  a  very  special  one. 

3.  Theorem.     If  a  is  a  root  of  the  equation  f(x)  =  0,  then  the 
quantic  f(x)  is  divisible  by  x—  a,  without  a  remainder. 


SOME    ELEMENTARY    PROPERTIES    OF   EQUATIONS         3 

Divide  the  polynomial  f(x)  by  x  —  a  until  a  remainder  is 
obtained  which  does  not  involve  x.  Designate  the  quotient 
by  Q,  the  remainder  by  R.  Then 


By  hypothesis,  a  is  a  root  ;  hence,  substituting  a  for  x,  we  have 


Consequently,  R  —  0,  and  the  theorem  is  proved.  The  follow- 
ing theorem  is  the  converse  of  this. 

4.  Theorem.     If  the  quantic  f(x)  is  divisible  by  x  —  a  without 
a  remainder,  then  a  is  a  root  of  f(x)  =  0. 

By  hypothesis,  f(x)  =  (x  —  a)  Q. 

The  equation  f(x)  =  0  may,  therefore,  be  written  (x  —  «)  Q  =  0, 
and  the  latter  is  seen  to  be  satisfied  when  a  is  substituted  for  x. 
Hence  a  is  a  root  of  f(x)  =  0. 

5.  The  preceding  theorem  is  a  special  case  of  the  following 

Theorem.  Ttie  value  of  the  quantic  /(»),  when  h  is  substituted 
for  x,  is  equal  to  the  remainder  which  does  not  involve  x,  obtained 
in  the  operation  of  dividing  f(x)  by  x  —  h. 

Let  R  be  the  remainder  which  does  not  involve  x  ;  then 


Substitute  h  for  x  and  we  obtain  f(Ji)  =  R. 

6.  Divisions  of  polynomials  by  binomials,  with  numerical 
coefficients,  may  be  performed  expeditiously  by  the  process 
called  synthetic  division.  Suppose  3?  +  5  x2  -f  4  x  —  23  is  to  be 
divided  by  x  —  3.  We  exhibit  the  ordinary  process,  and  also 
that  of  synthetic  division. 


THEORY   OF    EQUATIONS 


x-3  1  +  5  +    4-23[3_ 


^  +  80;  +  28  +3  +  244-84 

4x  1  +  8  +  28  +  61 

8<r-24o; 


28  a; -84 
61 

We  notice  that  in  synthetic  division  the  coefficients  are 
detached,  the  first  term  of  each  partial  product  is  omitted,  the 
second  term  of  the  divisor  has  its  sign  changed  so  that  the  sec- 
ond term  of  each  partial  product  may  be  added  to  the  correspond- 
ing term  of  the  dividend.  Moreover,  the  process  is  compressed 
so  that  the  coefficients  of  the  quotient  and  the  remainder  appear 
all  in  the  same  line. 

The  process  is  as  follows : 

Multiply  1  by  3  and  add  the  product  to  5,  giving  8. 

Multiply  8  by  3  and  add  the  product  to  4,  giving  28. 

Multiply  28  by  3  and  add  the  product  to  —  23,  giving  61. 

The  quotient  is  x1  +  8  x  +  28 ;  the  remainder  is  61. 

If  in  the  dividend  any  powers  of  x  are  missing,  their  places 
are  to  be  supplied  by  zero  coefficients. 

Divide  y?  —  2  $  +  x  —  5  by  x  +  5. 

1  +  0-   2+     0+     1-       51-5 
_  5  +  25  - 115  +  575  -  2880| 

1  _  5  +  23  -  115  +576  -  2885 

Hence  the  quotient  is  x*  —  5  a3  +  23  a?  —  115  x  +  576 ;  the 
remainder  is  —2885. 

Ex.  1.   Show  that  x4  —  Sx8  —  3x+15  has  5  as  a  root. 

1  _  5  +  o  -  3  +  15  15 
.  +  6  +  0  +  0_15|~ 

0+0-3+    0 
The  remainder  is  0  ;  hence,  by  §  4,  5  is  a  root. 

Ex.  2.   Show  that  x5  -  x4  +  10  x3  -  9  x2  +  8  x  +  699  =  0  is  satisfied 

by  x  =  -  3. 

.-    :- 

•  y 


Ex.  3.    Divide  x~  -  101  x5  +  x4  -  60  x2  +  x  by  x  +  4. 

Ex.  4.    If  /(x)  =  x5  -  6  x*  +  7  x8  +  x2  +  *  +  2,  find  the  value  of  /(10). 

Ex.  5.  Determine  the  value  of  the  quantic  x7  —  3x6+4x44-5x3  +  ll, 
when  x  =  —  6. 

Ex.  6.  If  -  4  is  a  root  of  2  x8  +  6  x2  +  7  x  -f  60  =  0,  find  the  other 
roots. 

Ex.  7.  Show  that,  if  /(x)  is  divided  by  x  —  ft,  each  successive  remain- 
der is  equal  to  /(A)  ,  when  h  is  substituted,  throughout,  for  x. 

7.  Theorem.  Every  equation  f(x)  =  0  of  the  nth  degree  has  n 
roots,  and  no  more. 

We  assume  here  that  every  such  equation  has  at  least  one 
root.  Let  «!  be  a  root  of  f(x)  =  0.  Then  f(x)  is  divisible  by 
x  —  a-i  without  remainder,  §  3  ;  so  that 

f(x)  =  (x-aj  ^(x), 

where  the  quotient  4>i(x)  *s  a  rational  integral  algebraic  func- 
tion of  x  of  the  (n  —  l)th  degree. 

Again  <£,(£)  =  0  has  a  root.  Denote  it  by  a%  then  <£i(#)  is 
divisible  by  x  —  a^  without  remainder,  so  that 


and  /(  x)  =  (x  —  a^)(x  —  a2 

Now  <£2(#)  is  a  rational  integral  algebraic  function  of  x  of 
the  (n  —  2)th  degree  ;  hence  <ft2(x)  =  0  has  a  root.  By  continu- 
ing in  this  way  we  shall  obtain  n  factors  of  f(x),  viz.,  x  —  «1? 
x  —  «2,  •••  x  —  «„,  and  the  only  other  factor  is  OQ,  which  is  the 
coefficient  of  xn  in  the  quantic  f(x).  Thus, 

f(x)  =  a0(x  -  «!)(*  -«,)...(«-  «„). 

As  the  quantic  /(#)  vanishes  when  we  put  for  x  any  one 
of  the  n  numbers  ccj,  a2,  '••  ctn,  it  follows  that  /(#)  =  0  has 
w  roots.  If  #  is  assigned  a  value  different  from  any  one 
of  these  n  roots,  then  no  factor  of  /(#)  can  vanish  and  the 
equation  is  not  satisfied.  Hence  f(x)  =  0  cannot  have  more 
than  n  roots. 


6  THEORY   OF   EQUATIONS 

8.  Theorem.  If  the  coefficients  of  f(x)  =  0  are  all  real,  then 
complex  roots  enter  the  equation  in  pairs. 

Let  a  -f-  ib  be  a  root  of  an  equation  f(x)  =  0  with  real  coef 
eients,  where  i=  V—  1  and  where  a,  but  not  b,  may  be  zero. 
We  shall  prove  that  the  conjugate  number,  a  —  ib,  is  also  a 
root. 

Substitute  the  root  a  +  ib  for  x  in  the  given  equation.  Then 
expand  the  powers  of  a  -f-  ib  by  the  binomial  theorem,  and 
simplify.  All  the  terms  which  do  not  contain  i  or  which  con- 
tain even  powers  of  i  will  be  real  ;  all  terms  which  contain  odd 
powers  of  i  will  be  imaginary.  Denote  the  algebraic  sum  of 
all  real  terms  by  P,  and  the  algebraic  sum  of  all  imaginary 
terms  by  iQ.  Then  we  have, 


But  this  equation  can  be  true  only  when  P=0  and  Q  =  0;  for 
the  real  and  imaginary  parts  can  never  destroy  each  other. 

Now  substitute  a  —  ib  for  x  in  the  equation  f(x)  =  0.  As 
before,  expand  and  simplify.  All  the  real  terms  will  be  un- 
changed ;  all  the  imaginary  terms  will  have  their  signs  changed, 
but  otherwise  will  be  the  same  as  before.  Hence  the  quantic 
/(a?)  now  assumes  the  value  P—  iQ.  But  we  have  shown  that 
P=  0  and  Q  =  0,  hence, 

p-;<3  =  o, 

that  is,  the  equation  f(x)  =  0  is  satisfied  by  x  =  a  —  ib.     Hence 
a  —  ib  is  a  root. 

9.  From  the  preceding  theorem  it  is  evident  that  every 
equation  of  odd  degree  and  with  real  coefficients  must  have  at 
least  one  real  root.  Thus,  a  cubic  equation  must  have  either 
three  real  roots  or  one  real  root  and  two  complex  roots. 

The  equation  x3  —  1  =0  has  evidently  the  real  root  1.  Divid- 
ing x3  —  1  by  x  —  1,  we  are  led  to  the  quadratic  x2  +  x  -f  1  =  0, 
both  roots  of  which  are  complex.  They  are  ^  —  1  ±  V  —  3  j  . 
The  three  roots  are  called  the  cube  roots  of  unity.  Observe  that 


.  V  ,  . 
,    ;  --^  f.    .  x    f.t 

SOME   ELEMENTARY    PROPERTIES   OF   EQUATIONS          7 

•    jf~^vJi.    ' 

the  square  of  either  complex  root  is  equal  to  the  other  complex 
root.  Also,  the  sum  of  the  three  roots  of  unity  is  zero. 

•|V10.  An  equation  f(x)  =  0  is  called  complete  when  all  the 
powers  of  x  from  xn  to  x°  are  present.  An  incomplete  equation 
can  be  made  complete  in  form  by  writing  the  missing  terms 
with  zero  coefficients. 

When  two  successive  terms  in  a  polynomial  or  in  an  equation 
have  the  same  sign,  there  exists  a  permanence  of  sign  ;  when 
1  two  successive  terms  have  opposite  signs,  there  exists  a  varia- 
tion of  sign.  In  the  equation  x5  -\-  x3  —  x2  +  5  =  0  the  signs 
occur  in  the  order  +  -\  ---  \-  and  there  are  two  variations.  and 
one  permanence. 

11.    Descartes'  Rule  of  Signs.     An  equation  f(x)  =  0,  the  coef- 
ficients of  which  are  real,  has  as  many  positive  roots  as  it  has 
^  _  ,  variations  of  sign,  or  fewer  by  an  even  number. 

o 

We  shall  show  that  if  a  polynomial  f(x)  is  multiplied  by  a 

factor  x  —  a,  thereby  introducing  a  new  positive  root,  the  varia- 
tions of  sign  in  the  product  will  exceed  those  in  the  polynomial 
,--•   by  an  odd  number. 

In  the  function  /(a?),  which  is  arranged  according  to  the  de- 
,x   scending  powers  of  x  and  may  be  either  complete  or  incomplete, 
•'^  we  assume  that  the  signs  of  the  terms  vary  in  the  following 
manner  :  , 

•  T  "•> 

where  the  dots  which  follow  a  -f  stand  for  any  given  number 
of  consecutive  terms  which  are  positive  and  where  the  dots  which 
follow  a  —  designate  consecutive  terms  which  are  negative. 

Let  a  be  a  positive  root.  Multiplying  f(x)  by  x  —  «,  and 
writing  like  powers  of  x  underneath  each  other,  we  obtain  a 
product  whose  signs  may  be  written  as  follows: 


+  ±  ----  ±-  +  ±  ----  ± 


8  THEORY   OF    EQUATIONS 

The  ±  denotes  an  ambiguity;  that  is,  the  sign  of  a  term  so 
affected  is  here  undetermined.  We  see  that  the  dots  which 
follow  ±  are  ambiguities ;  that  is,  each  permanence  of  sign  in 
f(x)  is  here  replaced  in  (x  —  «)  •/(#)  by  an  ambiguity.  We 
see  also  that  to  every  variation  of  sign  in  f(x)  there  corresponds 
a  variation  in  (x  —  a)  •/(#)•  In  the  product  there  is,  in  addi- 
tion, a  variation  introduced  at  the  end.  Hence  the  product 
contains  at  least  one  more  variation  than  does  f(x).  It  may 
contain  more ;  for,  successive  permanences  like  -\ — | — |-  or 
,  occurring  in  f(x)  and  replaced  in  (x  —  «)  •/(#)  by  ambi- 
guities, may  in  reality  be  replaced  by  the  signs  +  ; — f-  or 
— | — .  But  such  changes  in  sign  ajways  increase  the  varia- 
tions by  an  even  number.  Hence  in  (x  —  a)  •  f(x)  the  total  num- 
ber of  variations  exceeds  that  in  f(x)  by  the  odd  number  1  or 
1  +  2  li. 

The  same  conclusion  is  reached  when  the  last  term  in  f(x) 
is  negative. 

Descartes'  Riile  follows  now  easily.  Suppose  the  product  of 
all  the  factors,  corresponding  to  negative  and  complex  roots  of 
f(x)  =  0,  to  be  already  formed.  Designate  this  product  by 
F(x).  Since  F(x)  =  0  has  no  positive  roots,  the  first  and  last 
terms  in  F(x)  have  like  signs.  Hence  the  number  of  varia- 
tions in  F(x)  is  an  even  number,  2  k,  where  k  is  zero  or  a  posi- 
tive integer.  Now,  if  F(x)  is  multiplied  by  the  factor  x  —  a1} 
where  «t  is  a  positive  root,  we  get  in  the  product  2  7^  +  1  vari- 
ations, where  A^  ^>  k.  In  the  same  way  a  second  factor  x  —  u.2 
gives  rise  to  2  k2  +  2  variations,  and  so  on.  Thus,  the  intro- 
duction of  v  positive  roots  results  in  2  kv  +  v  variations,  where 
A1,,  is  zero  or  a  positive  integer.  Hence,  the  theorem  is 
established. 

12.  Negative  Roots.  To  apply  Descartes'  Rule  to  negative 
roots  of  f(x)  =  0  we  write  down  an  equation  whose  roots  are 
those  of /(#)  =  0  with  their  signs  changed.  The  new  equation 
can  be  derived  by  substituting  in  f(x)  =  0,  —  x  for  x.  The 


SOME  ELEMENTARY  PROPERTIES  OF  EQUATIONS    9 

process  merely  alters  the  signs  of  all  the  terms  involving  odd 
powers  of  x.  It  is  readily  seen  that  if  a  satisfies  the  equation 
f(x)=0,  then  —  a  satisfies  the  equation  f(—x)  =  0.  Hence, 
each  negative  root  of  f(x)  =  0,  with  its  sign  changed,  is  a  posi- 
tive root  of  /(—  #)  =  0.  Descartes'  Rule  may  now  be  applied 
to /(— »)  =  (). 

Ex.  1.   Determine  the  nature  of  the  roots  of  a;3  +  3  x  +  1  =  0. 

There  is  no  variation  ;  therefore,  no  positive  root.  Transform  the  equa- 
tion by  changing  the  signs  of  the  terms  containing  odd  powers  of  x.  We 
get  Xs  +  3  x  —  1  =  0.  The  new  equation  has  one  variation  ;  hence, 
cannot  have  more  than  one  positive  root.  Consequently,  the  original 
equation  cannot  have  more  than  one  negative  root.  The  real  root  of  the 
given  cubic  is  thus  seen  to  be  negative ;  the  other  two  roots  must  be 
complex. 

Ex.  2.  Apply  Descartes'  Rule  to  f(x)  =  x*  -  Xs  +  1  x  +  6  =  0.  Here 
/(x)  has  two  variations,  and  /(  —  x)  has  two  variations.  Hence  /(x)  =  0 
cannot  have  more  than  two  positive  roots  nor  more  than  two  negative 
roots. 

Ex.  3.  Apply  Descartes'  Rule  to  xZn  —  1=0.  Since  x2™  —  1  has  one 
variation  and  (—  x)2"  —  1  has  one  variation,  the  given  equation  cannot 
have  more  than  one  positive  root  nor  more  than  one  negative  root.  We 
readily  see  that  +  1  and  —  1  are  roots.  Hence  there  are  2  n  —  2  complex 
roots. 

Ex.  4.  Prove  that  if  the  roots  of  a  complete  equation  are  all  real,  the 
number  of  positive  roots  is  equal  to  the  number  of  variations,  and  the 
number  of  negative  roots  is  equal  to  the  number  of  permanences. 

Ex.  5.  An  equation  with  only  positive  terms  cannot  have  a  positive 
root.  If  the  number  of  variations  is  odd,  the  equation  has  at  least  one 
positive  root,  but  it  cannot  have  an  even  number  of  positive  roots. 

Ex.  6.  A  complete  equation  with  alternating  signs  cannot  have  a 
negative  root. 

Ex.  7.  If  all  the  terms  of  an  equation  are  positive  and  the  equation 
involves  no  odd  powers  of  x,  then  all  its  roots  are  complex. 

Ex.  8.  If  all  the  terms  of  an  equation  are  positive  and  all  involve  odd 
powers  of  x,  then  0  is  the  only  real  root  of  the  equation. 


10  THEORY   OF   EQUATIONS 

Ex.  9.    Apply  Descartes'  Rule  to 

xs  -  21  x  +  20  =  0.  yf>  -f  of  +  1  =  0. 

x3-x2  +  10x-15  =  0.  x6-l  =  0. 

x*-t-5x3-4x2-3x  +  5  =  0.  z8-x*  +  x2  +  1  =0. 

x4  +  1  =  0.  xn  +  1  =  0. 

xfi  +  l^O.  x»-l=0. 

x5  -  1  =  0. 

Ex.  10.     The  equation  x4  -  4  x3  -  7  x2  +  22  x  +  24  =  0  has  no  com- 
plex roots.     How  many  are  positive  ?     How  many  are  negative  ? 

Ex.  11.     Show   that  x5  —  x4  +  x3  —  x2  —  x  —  1  =  0  cannot  have  just 
two  positive  roots  nor  just  one  negative  root, 

<—  ^  "  "1  W» 

13.   Relations  between  Roots  and  Coefficients. 

If  /(a?)  =  zn  +  a^"-1  +  •  •  •  +  a»-ix  +  an  =  0 

has  the  roots  «15  a2,  •••,  «„,  then,  by  §  7,  we  have 
/(a?)  =  (x  -  «,)(*  -  «a)  ...  (a;  —  «„)  =  0. 

If  w  be  taken  successively  equal  to  2,  3,  or  4,  we  obtain  by 
ordinary  multiplication, 


f(x)  =  (x  —  a^)(x  —  a2)  =  x2—  («!  +  a2}x  +  «1«2  =  0, 
f(x)  =  (x  —  «j)  (a;  — 


-  «2)(a;  ~  "sX*  -  *«4)  =  ^  -  («1  +  «2  +  «3  +  «4) 


«i«3«4  4-  «2«3«4)#  -)-  «1a2a3a4  =  0. 
These  relations  are  seen  to  obey  the  following  laws  : 

In  the  equation  f(x)  =  0,  in  which  the  coefficient  of  xn  is  unity, 
the  coefficient  at  of  the  second  term,  with  its  sign  changed,  is  equal 
to  the  sum  of  the  roots. 


SOME  ELEMENTARY  PROPERTIES  OP  EQUATIONS   11 

Tlie  coefficient  a.2  of  the  third  term  is  equal  to  the  sum  of  the 
'products  of  the  roots  taken  two  by  two. 

TJie  coefficient  a3  of  the  fourth  term,  with  its  sign  changed,  is 
equal  to  the  sum  of  the  products  of  the  roots  taken  three  by  three; 
•and  so  on,  the  signs  of  the  coefficients  being  taken  alternately 
negative  and  positive,  and  the  number  of  roots  taken  in  each  iirod- 
uct  increasing  by  unity  every  time  we  advance  to  a  new  coef- 
ficient, until  finally  the  last  term  in  the  equation  is  reached, 
which  is  numerically  equal  to  the  product  of  all  the  roots  and 
which  is  positive  or  negative  according  as  n,  the  degree  of  the 
equation,  is  even  or  odd.  In  symbols,  these  laws  may  be  ex- 
pressed as  follows : 


an=    — 

When  in  the  equation  f(x)  =  0  the  coefficient  a0  of  the  term 
xn  is  not  unity,  we  must  divide  each  term  of  the  equation  by  a0. 

The  .sum  of  the  roots  is  then  equal  to -,  the  sum  of  their 

fl  ^0 

products,  two  by  two,  is  — ,  and  so  on. 
a0 

The  laws  expressing  the  relations  between  the  coefficients 
of  an  equation  and  the  roots  were  obtained  above  by  observ- 
ing the  relations  existing  in  the  three  products  obtained  by 
actual  multiplication.  To  remove  any  doubt  which  may  be 
entertained  as  to  the  generality  of  these  laws  Ave  proceed  as 
follows.  Suppose  these  laws  to  hold  when  n  factors  are  multi- 
plied together ;  that  is,  suppose  that 

(x  —  a^(x  —  «2)  •••  (x  —  «„)  = 
where  a1}a2,  •••,a 

have  the  values  shown  in  I. 


12  THEORY   OF   EQUATIONS 

Multiply  both  sides  of  this  identity  by  another  factor  x  —  «M+1, 
and  we  get 


(a;  -  «!)(*  -  «2)  •••  (a;  -  «„)(*  -  ccn+1)  =  ccn+1  +  («!  -  «n+1)a;B 
+  (a2  —  Ci^+Oa?"-1  +  ...  —  anan+1. 

But  ttj  —  «n+1  =  —(«!  +  «2  +    •  •  *    +  «n)  —  «»+!> 

•    +  «»-!«»)  4- 


—  a2«n+1  =  — 


«  an+1. 

Hence,  if  the  laws  hold  for  n  factors,  they  hold  for  n  +  1 
factors.  But  from  actual  multiplication  we  know  that  the 
laws  hold  when  n  =  4,  therefore  they  must  hold  when  n  =  5. 
Holding  for  n  =  5,  they  must  hold  when  n  =  6,  and  so  on  for 
any  positive  integral  value  of  n. 

14.  It  might  appear  that  the  n  distinct  relations  existing 
between  the  coefficients  and  roots  of  an  equation  of  the  nth 
degree  should  offer  some  advantage  in  the  general  solution  of 
the  equation,  that  one  of  the-n  roots  could  be  obtained  by  the 
elimination  of  the  (n  —  1)  roots  from  the  n  equations.  But 
this  process  offers  no  advantage,  for  on  performing  this  elimi- 
nation we  merely  reproduce  the  proposed  equation.  Take,  for 
example,  the  cubic  v?  -f  a^2  +  a^  -j-  as  =  0. 

We  have  «1  =  -  «  -  «-  « 

cr2  = 


To  eliminate  «2  and  «3,  multiply  both  sides  of  the  first  equa- 
tion by  «!2,  both  sides  of  the  second  by  al9  and  add  the  results 
to  the  third  equation. 

We  obtain  «,»  +  o^  +  a^  +  a,  =  0, 


SOME  ELEMENTARY  PROPERTIES  OF  EQUATIONS   13 

which  is  simply  the  old  equation  with  a^  in  place  of  x  to  repre 
sent  the  unknown  quantity.  ' 

While  the  equations  expressing  the  relations  between  roots 
and  coefficients  offer  no  advantage  in  the  general  solution  of 
equations,  they  are  of  service  in  the  solution  of  numerical 
equations  when  some  special  relation  is  known  to  exist  among 
the  roots.  Moreover,  in  any  algebraic  equation  they  enable  us 
to  determine  the  relations  between  the  coefficients  which  corre- 
spond to  some  given  relations  between  the  roots. 

Ex.  1.  The  cubic  x3  +  3  x2  —  16  x  —  48  =  0  has  two  roots  whose  sum 
is  zero.  Solve  the  equation. 

We  have  a\  +  «2  =  0, 

«i  +  «2  +  «s  =  —  3. 

Hence  «3  =  —  3.     Dividing  the  cubic  by  x  +  3,  we  have 
x2  -  16  =  0,  x  =  ±  4. 

Ex.  2.  The  roots  of  the  cubic  x9  —  9  x2  +  26  x  —  24  =  0  are  in  arith- 
metical progression.  Find  them. 

Let  a  —  d,  a,  a  +  d  be  the  three  roots. 

Then  3  a  =  9,  3  a2  -  d2  =  26  ;  therefore  a  =  3,  d  =  1,  a  -  d  =  2, 
a  +  d  =  4.  The  roots  are  2,  3,  4. 

Ex.  3.  Two  roots  of  the  cubic  3  x3  +  x2  —  15  x  —  5  =  0  have  the  sum 
zero.  Find  all  three  roots. 

Ex.  4.  The  equation  2x3  +  7x2  +  4x  —  3  =  0  has  two  roots  whose 
sum  is  —2.  Solve  the  equation. 

Ex.  5.  The  equation  2  xs  +  23  x2  +  80  x  +  75  =  0  has  two  equal  roots. 
Solve. 

Ex.  6.  The  biquadratic  equation  9  x4  +  42  x3  +  13  x2  -  84  x  +  36  =  0 
has  two  pairs  of  equal  roots.  Find  them. 

Ex.  7.  If  the  equation  x4  +  Oix3  +  a2x2  +  «3x  +  a4  =  0  has  all  its  roots 
equal,  what  relation  exists  between  its  coefficients  ? 

Ex.  8.  Show  that  the  sum  of  the  nth  roots  of  unity  is  zero. 

15.  Symmetric  Functions.  If  a  function  of  two  or  more 
quantities  is  not  altered  when  any  two  of  the  quantities  are 


14  THEORY   OF   EQUATIONS 

interchanged,  it  is  called  a  symmetric  function.  For  example, 
the  trinomial  cr-\-l>-  +  c-  is  a  symmetric  function  of  a,  b,  c, 
because,  if  any  two  quantities,  say  a  and  6,  are  interchanged, 
the  expression  is  unaltered  in  value.  We  are  concerned  mainly 
with  symmetric  functions  of  the  roots  of  an  equation.  The 
simplest  examples  of  such  functions  are  those  given  in  §  13, 

viz.. 

«i  +  «2+«3+  '••  +  «„, 

«2«3  H  -----  \-  «n-l«n, 

h  a»_2«»-i«n    etc. 


These  are  the  simplest,  because  in  no  term  does  any  one  of 
the  roots  occur  to  a  higher  power  than  the  first.  Other  ex- 
amples of  symmetric  functions  of  the  roots  are 


We  shall  represent  a  symmetric  function  by  the  letter  2^ 
followed  by  one  of  the  terms  of  the  function.  Given  the  roots 
and  one  of  the  terms  of  the  symmetric  function  of  these  roots, 
it  is  usually  not  difficult  to  write  down  all  the  terms  of  the 
function.  Thus,  given  the  roots  a,  ft,  y  of  a  cubic  equation,  then- 

2«  =  a  +  ft  +  y, 
=  a/3  +  ay  +  (3y, 
«2y 


*  Ex.  1.   If  x3  +  ax2  +  bx,  +  c  =  0  has  the  roots  a,  /3,  7,  express  the  value 
of  2«20  in  terms  of  the  coefficients. 

Multiply  «  +  /3  +  7  =  _a 

by  «)3  +  ay  +  0y  =  b, 

and  we  obtain  S«20  +  3  afiy  —  —  ab 

and  2«2y3  =  3c  —  ab. 

Ex.  2.   Find  S«2  for  the  same  cubic. 

*The  results  in  an  example  marked  with  a*  will  be  used  in  later  examples. 


SOME   ELEMENTARY   PROPERTIES   OF   EQUATIONS      15 

*  Ex.  3.    Find  S«3  for  the  same  cubic. 

Multiply  the  functions  Sa  and   2«2  together,   and  the  product   is 
S«3  +  S«2/3.     Hence  S«3  =  Sa  •  S«2  -  S«2/3  =  -  a3  +  \ab  -  3  c. 


Ex.  4.   For  the  same  cubic,  find 

Squaring  both  sides  of  «/3  +  #7  +  py  =  6,  we  obtain 

a?p  +  «272  +  jSV  +  2  «/37(«  +  /3  +  7)  =  52. 

Ex.  5.   For  the  same  cubic,  find  Sa3/}. 

Show  that  S«2|8  •  Set  =  S«3/3  +  2  Sa2^  +  2  apy(a  +  /3  +  7). 


Ex.  6.   For  the  same  cubic,  find  the  value  of  («  +  /3)(/3  +  7)  (7  +  a). 

Ex.  7.   If  x4  +  ax3  +  foe2  +  ex  +  d  =  0  has  the  roots  a,  j8,  7,  5,  find  the 
value  of  S«2. 

Ex.  8.   For  the  same  quartic,  find  the  value  of  S«2/3. 
Ex.  9.   For  the  same  quartic,  find  the  value  of 


Ex.  10.    Find  the  value,  expressed  in  terms  of  the  coefficients,  of  the 
sum  of  the  squares  of  the  roots  «j,  «2,  •••,  «n,  of 


xn  +  aix"-1  +  a2x"-2  +  •••  +  an-ix  +  an  =  0. 
Squaring  S«j  =  —  ai,  we  get  'Za^  +  2  2«i«2  =  ai2,  hence 
S«i2  =  ai2  -  2  «2. 

Ex.  11.  In  the  same  equation,  find  ?  —  • 
By  §  13  we  have  "^  Kl 

'  On  H 


Dividing  the  former  by  the  latter  we  obtain 


«1        «2        «3  On 

Ex.  12.   Find  the  sum  of  the  reciprocals  of  the  roots  of  the  equation 
s6  +  x2  +  10  x  -f  105  =  0.     Find  also  V— 

•^«« 


16.  Graphic  Representation  of  the  Polynomial  f(x).  The 
changes  in  value  of  the  polynomial  f(x)  =a0an-f-a1aJn~1H  -----  \-an, 
as  the  variable  x  increases  or  decreases,  can  be  seen  most  easily 
by  the  aid  of  graphic  representations. 


16 


THEORY   OF   EQUATIONS 


Let  XX'  and  YY'  be  two  perpendicular  lines,  called  axes  of 
reference.  Their  intersection  0  is  called  the  origin.  Let  values 

of  x  be  measured  off  from  the  origin 
O  along  the  axis  XX'  and  values  of 
y  be  measured  off  from  O  along  the 
axis  YY'.  Positive  values  of  x  are 
measured  from  0  toward  the  right ; 
negative  values,  toward  the  left. 
Positive  values  of  y  are  measured 
from  0  upward ;  negative  values  of 
y,  downward. 

The  distances  of  a  point  P  from  the  axes  of  reference  are 

called  the  coordinates  of  the  point.     Thus,  Pm  and  Pn  are  the 

coordinates  of  the  point  P,  both  coordinates  being  positive;  Q£ 

and  Qrare  the  coordinates  of  the  point  Q,  both  being  negative. 

Let  y  represent  the  value  of  the  polynomial  f(x)  ;  that  is,  let 

2f  =/(*)•    < 
\?  4* 

Suppose  now  that  y  =  Pn  when  x  =  Pm,  then  the  position  of  the 

point  P  represents  to  the  eye  simultaneously  the  value  of  x  and 
the  corresponding  value  of  /(#).  If  different  values  of  x  be  laid 
off  on  the  axis  XX '  and  the  corresponding  values  of  f(x)  on 
ihe  axis  YY',  the  points  thus  located  will  all  lie  on  a  line  or 
jurve,  called  the  graph  of  the  polynomial  /(#). 

In  the  construction  of  the  graphs  of  polynomials  it  is  con- 
venient to  use  "plotting"  or  "coordinate"  paper,  ruled  in 
small  squares. 

Ex.  1.   Construct  the  graph  of /(x)  =  x2  +  x  —  2. 

Putting  y  =  x2  +  x  —  2,  we  readily  compute  the  following  sets  of  values : 


If 


x  =  0,      y  =  -  2. 

*=±i,  y  =  -ll  or  -2J. 

a;=±  1,  y  =  0  or  --2. 

x  =  ±  2,  y  =  4  or       0. 

x  =  ±  3,  y  =  10  or      4. 


SOME   ELEMENTARY   PROPERTIES    OF   EQUATIONS      17 


Plotting  these  points  we  get  the  ad- 
joined curve.  Here  unity  is  taken  equal 
to  |  of  a  side  of  a  square. 

From  the  shape  of  this  curve  we  can 
see  that  when  x  is  negative  and  increases, 
then  /(a;)  decreases  and  reaches  a  minimum 
value  when  x=  —  \.  From  there  on,  as  x 
increases,  the  /(x)  increases.  The  curve 
is  a  parabola.  It  cuts  the  axis  XX'  in 
two  places;  that  is,  there  are  two  values 
of  X,  for  which  the  value  of  /(x)  is  zero. 
These  two  values  of  x  are  1  and  —  2. 
Hence  1  and  —  2  are  roots  of  the  equa- 
tion /(x)  =  0. 

Ex.  2.   Construct  the  graph  of  f(x)  =  \  xz  +  x  +  3. 
If  x  =  0,       y  =  3. 


x  =  ±2,y  =  Gl    or  2J. 
x  =  ±  3,  y  =  9      or  3. 
x  =  ±  4,  y  =  12£  or  4J. 


-X- 


The  curve  does  not  cut  the  axis  XX' ;  hence  no  real  value  of  x  makes 
f(x)  zero,  and  the  roots  are  both  imaginary, 
c 


18 


THEORY    OF    EQUATIONS 


Ex.  3.    Construct  the  graph  f(x)  -  x3  -  z2  +  2  x  -  3. 
i      x  =  0,       y  =  -  3. 

x  =  ±  .5,  y  =  —  2.12  or  —  4.37. 
s  =  ±  1,  y  =  —  1      or  —  7. 
a:  =  ±  2,  j/  =  5          or  -  19. 
x  =  ±3,  y  =  21        or  —  45. 


The  curve  crosses  the  axis  A' A"'  only  once  ; 
hence  there  is  only  one  real  root.  The  value 
of  this  root  is  seen  from  the  figure  to  be 
about  1.3. 

Ex.  4.    Find  the  graph  of  z3  +  a;2  +  2  x  -  4. 
Ex.  5.  Find  the  graph  of  z4  —  2  x  +  1. 

17.  In  constructing  the  graph  of  a 
polynomial  f(x)  we  located  a  num- 
ber of  points  and  then  drew  a  curve 
through  them.  The  curve  thus  ob- 
tained was  assumed  to  represent  the 
continuous  variation  of  the  value  of 
f(x),  corresponding  to  the  continuous 
increase  of  x.  But  this  assumption 
that  the  polynomial  /(#)  never  jumps 
from  one  value  to  another,  when  x  is  made  to  vary  continuously 
from  one  value  to  another,  requires  proof.  The  proof  will  be 
given  in  §  25.  It  is  facilitated  by  the  use  of  derived  functions 
and  Taylors  TJieorem. 

18.   Derived  Functions  and  Taylor's  Theorem.     In 

/(»)  =  OoZ"  +  a^"-1  -f  a^"-2  H h  CLn^x  +an, 

let  x  receive  an  increment  h  and  write  x  +  h  in  place  of  x.    We 
have 
f(x  +  K)  =  a0(x  +  h)n  +  a^x  +  /i)"-1  +  •  •  •  +  «n_i  0  +  A)  +  an. 

Let  each  term  be  expanded  by  the  binomial  formula.     Then 
collect  the  coefficients  of  like  powers  of  h,  and  we  get 


SOME   ELEMENTARY   PROPERTIES    OF    EQUATIONS      19 

.f(x  +  h)=  aoz"  +  a^"-1  +  a^"-2  +  -.  +  <*„_,»  +'an 
+h\na<{cn-1  +  0  -  l/aX1-2*  (n  -  2)a^"-3  +.—  +  a^f 


The  first  line  in  this  expansion  is  obviously  /(a:)  .  We  shall 
call  the  coefficient  of  h  the^rsf  derived  function  and  denote  it 

h2 

by  f(x).      Similarly  we  shall  call  the  coefficient  of  -  the 

1  •  2i 

second  derived  function  and  denote  it  by  f"(x);  and  so  on. 
The  rth  derived  function  is  designated  by  /'"(#)•  In  ^ne  Dif- 
ferential Calculus  these  derived  functions  are  called  differential 
coefficients.  Using  this  new  notation,  the,  above  result  may  be 
written  as  follows  : 


In  the  Differential  Calculus  this  scries  goes  by  the  name  of 
Taylor's  TJieorem.  We  have  here  established  the  truth  of  this 
theorem  for  rational  integral  functions  of  x,  but  the  theorem 
has  actually  a  much  wider  application. 

The  results  of  this  paragraph  are  true  of  complex  numbers, 
as  well  as  of  real  numbers. 

19.  To  arrive  at  a  convenient  rule  for  finding  derived  func- 
tions, compare  the  following  expressions  : 


f(x)  =  a^Kn  +  a^-1  +  a^"-2  -|  -----  h  «„-!»  4-  «„, 
f(x)  =  na^cn~l  +  (n-  1)^*-2  +  (n  -  2)a2a;"-3  +  . 
/"(a?)  =  n(n  -  1)^"-2  +  (n  -  1)  (n  -  2)a1xn~*-\  -----  (-2  an 


20  THEORY   OF   EQUATIONS 

We  observe  that/'  (x)  can  be  obtained  from  f(x)  in  this  man- 
ner :  Multiply  each  term  inf(x)  by  the  exponent  of  x  in  that  term, 
and  diminish  the  exponent  of  x  in  the  term  by  unity.  By  this 
rule  a<pn  becomes  wa^r""1,  etc.  ;  an,  i.e.  anx°,  becomes  0  •  anx~l, 
or  0.  Notice  that  f"(x)  can  be  derived  from  f(x)  in  the  same 
way  as  f'(x)  was  derived  from/(x). 

Ex.  1.   If  /(x)  =  x5  +  8  x*  +  5  xs  +  6  z2  +  7  x  +  10, 
then  /'(x)  =  5z4  +  12x3+15z2+12z  +  7, 

/"(x)  =  20  x3  +  36  x2  +  30  x  +  12, 


/"'(z)=120x  +  72, 
/v(x)=120. 

Ex.  2.  Find  all  the  derived  functions  of 

x6  +  2  x5  +  7  x3  +  8  x2  +  15. 

20.   Another  Form  of  f'(jf).     By  §  7, 

f(x)  =  a0(£  -  «0  (*—  02)  (a;  -  «3)  •••  (a  —  «„). 
Letting  #  increase  to  x  +  ft,  we  have 

/(a;  +  ft)  =  a0(x  +  ft  —  ax)  (x  +  ft  —  02)  •  ••  (x  +h  —  an). 
But,  by  Taylor's  Theorem,  §  18, 


Hence  the  coefficient  of  h  is  /'(#),  and  /'(#)  must,  therefore, 
be  equal  to  the  coefficient  of  h  in  the  right  member  of  I. 

That  is,    /'(aj)=a0(aj-a8)(aj-<n3)".(a;—  an)+a0(a;-a1)(aj—  «3) 

...(iK_an)+...=  /(^  +  /(^_+...+  /^)_.         n 

x—cti     x—a2  x—(*-n 

Formula  II  is  still  true  if  some  of  the  roots  are  equal.     Sup 
pose  ttj  occurs  as  a  root  s  times  and  «2  occurs  t  times,  then 

f(x)=a<)(x-aly(x-a2)t—  , 
and  formula  II  becomes 


X  — 


SOME   ELEMENTARY   PROPERTIES   OF    EQUATIONS      21 

Ex.  1.    If  /(x)  =  (x  -  1)  (x  -  2)  (z  -  3),  show  that 

f'(x)  =  (x  -  2)  (x  -  3)  +  (x  -  1)  (x  -  3)  +  (x  -  1)  (z  -  2). 

Ex.  2.   If  /(x)  =  (as  -  l)3(x  -  2)2,  show  that 

/'(x)  =  3(x  -  l)2(x  -  2)2  +  2(x-l)8(x-  2). 

Ex.  3.   If  /(ar)  =  (a;  -  a)'(x  -  6)<(x  -  c)«,  show  that 

/'(x)  =s(x-a)-1(x-&)<(x  -  c)«+«(x  -  a)'(x  -  6)«-»(*  -  c)* 


Ex.  4.  If  /(x)  =  (x  —  «i)  (x  -  «2)  (£  —  «s)  =  0,  show  that 


21.  Multiple  Roots.  If  we  consider  the  general  equation  in 
the  factored  form 

(x  —  cti)  (x  — a?)  (x— <%)•••  (x  — 0^  =  0, 

it  is  evident  that,  in  special  cases,  two  or  more  factors  may  be 
equal  to  each  other,  yielding  equal  or  multiple  roots. 

Suppose  that  m  roots  are  equal  to  each  other ;  then  there  are 
ra  equal  factors,  and  /(a;)  may  be  written 

Then  f'(x)  =  m(x  —  a1)ra~1^>(a;)  +  (x  —  a^)m<tf(x), 

and/(#)  and  f'(x)  have  the  factor  (a;  —  a^™*1  in  common.  This 
fact  suggests  the  following  process  for  the  discovery  of  mul- 
tiple roots:  Find  the  highest^common  factor  between  f(x)  and 
f'(x).  Suppose  this  factor  is  (x  —  a^f,  then/fee)  has  the  factor 
(x  —  «j)r+1,  and  there  are  r  +  1  equal  roots.  That  is,  «t  occurs 
as  a  root  r  +  1  times.  Suppose  the  highest  common  factor  to 
be  (a;  —  al)r(x  —  «.,)',  then  ^  occurs  as  a  root  r  +  1  times  and  ay 
occurs  as  a  root  s  +  1  times. 

Ex.  1.    Examine  8  z8  -  20  x2  +  6  x  +  9  =  0  for  equal  roots. 

f'(x)  =  24  x2  -  40  x  +  6,  and  the  H.  C.  F.  of  /(x)  and  /'(x),  found  by 
U>e  process  of  successive  divisions,  is  2  x  —  3.  Hence  (2  x  —  3) 2  is  a  fac- 
tor of  /(x),  and  f  is  a  double  root.  The  adjoining  figure  is  the  graph  of 


22 


THEORY   OP   EQUATIONS 


/(x).     At  x  =  |  the  curve  is  touched  by  the  axis  XX1.     In  other  words, 
the  axis  is  tangent  to  the  curve  and  meets  it  in  two  coincident  points. 

These  reveal  graphically   the  presence  of  a 

double  root.     The  third  root  is  seen  from  the 
figure  to  be  x  =  —  \. 

If  the  entire  curve  were  moved  downward, 
both  axes  remaining  fixed,  then  the  axis  OX 
would  become  a  secant  line ;  instead  of  the 
two  coincident  points  we  would  have  two 
distinct  points  of  intersection,  and  the  two 
roots  would  be  unequal.  If  the  curve  were 
shifted  bodily  upward,  then  the  part  of  it  to 
the  right  of  the  axis  YY1  would  have  no  point 
in  common  with  the  axis  XX1,  and,  instead 
of  two  equal  roots,  we  would  have  two  com- 
plex roots.  Thus  equal  roots  are  seen  to  be 
a  connecting  link  between  distinct  real  roots 
and  complex  roots. 

Shifting    the    curve    upward    corresponds , 
to  increasing  the  value  of  the  absolute  term 
in  f(x) ;   shifting   it   downward  corresponds 
to    diminishing    the    value    of    the    absolute 
term. 

Ex.  2.  Tell  from  the  graph  in  Ex.  1,  §  16,  by  about  how  much  the 
absolute  term  in  f(x)  must  be  increased  to  yield  equal  roots ;  to  yield 
complex  roots. 

Ex.  3.  Find  the  multiple  roots  of  8  x4  +  20  x9  +  18  x2  +  7  x  +  1  =  0. 
The  H.  C.  F.  of /(a;)  and /'(x)  is  4x2  +  4x  -f  1  =  (2x  +  I)2;  hence  -£ 
is  a  triple  root.  Construct  the  graph  for /(a;). 

Ex.  4.    Find  the  multiple  roots  of 

4  x5  -  8  x4  -  23  xs  +  19  x?  +  55  x  +  25  =  0. 
Ex.  5.    Find  the  roots  and  construct  the  graph  of 

x5  -  3  x4  +  3  x3  -  x2  =  0. 

Ex.  6.    Find  the  equal  roots  and  construct  the  graph  of 
X4  _  e  x3  +  13  x2  -  12  x  +  4  =  0. 

Ex.  7.  Prove  that,  if  a  occurs  as  a  root  of  /(x)  =  0  m  times,  then  a 
satisfies  each  of  the  equations  /(x)  =  0,  /'(x)  =  0,  •••/m~1(x)  =  0. 


SOME   ELEMENTARY   PROPERTIES    OF   EQUATIONS      23 


22.  Graphic  Representation  of  Complex  Numbers.     In  the  con- 
struction of   graphs  of   polynomials   y=f(%)  we  assumed  a 
horizontal  and  a  vertical  axis,  and  from  this  point  of  inter- 
section measured  off  values  of  x  parallel  to  the  horizontal  axis 
and  values  of  y  parallel  to  the 

vertical  axis.  A  similar  plan  is 
commonly  adopted  for  the  repre- 
sentation of  complex  numbers 
or  imaginaries.  If  z  =  x  -f  iy, 
where  x  and  y  are  real  numbers, 
either  -f-  or  — ,  rational  or  irra- 
tional, then  x  and-?/  jire  laid  off 
parallel  to  the  horizontal  and 
vertical  axis,  respectively.  If 
x  =  OQ,  y  =  QP,  then  z  is  repre- 
sented in  magnitude  and  direction  by  OP.  The  length  of  OP 
igJ^3[j[g.(LJJle  modulus  of  z,  and  is  equal  to  -Vx2  +  y2.*  The 
direction  of  z  is  indicated  by  the  angle  0,  which  is  called  the 
amplitude  or  argument  of  z. 

Since  x  =  p  cos  6,  y  =  p  sin  6,  we  have 

z  =  x  -f-  iy  =  p(cos  0  +  i  sin  0). 

This  graphic  representation  of  complex  numbers  is  due  to 
Caspar  Wessel  (1797). 

23.  Addition  and  Subtraction  of  Complex  Numbers.     Let  OP 

=  a  +  ib  and  OP'  =  a'  +  ib',  then,  OP+OPl  =  (a+a')+i(b+b'~). 
DrawP'S  parallel  and  equal  to  OP,  then  OT=a+ a',  TS=b  +  b', 
and  OS=OP+OP'. 

*  This  graphic  representation  is  of  great  help  to  the  mathematician.  But 
attention  should  be  called  to  the  fact  that  the  statement,  that  to  every  irra- 
tional number  there  corresponds  a  line  of  definite  length ,  is  no  longer  con- 
sidered self-evident  nor  demonstrable;  it  involves  the  geometric  postulate: 
"  If  all  points  of  the  line  fall  into  two  classes  in  such  a  manner  that  each 
point  of  the  first  class  lies  to  the  left  of  each  point  of  the  second  class,  then 
there  exists  one  point,  and  only  one,  which  brings  about  this  separation." 
See  the  Eucyklopiidie  d.  Math.  Wiss.,  I A  3,  No.  4. 


24 


THEORY    OF    EQUATIONS 


Using  the  notation  p  =  mod.  OP,  we  readily  see  that,  in  this 


case, 


mod.  OS  <  mod.  OF  +  mod.  PS. 


This  means  simply  that  two  sides  of  a  triangle  are,  together, 
greater  than  the  third  side.  If  OP  and  OP'  had  the  same 

amplitude  (that  is,  the  same 
direction),  then  the  modulus 
of  their  sum  would  be  equal 
to  the  sum  of  their  moduli. 
Extending  these  considera- 
tions to  three  or  more  im- 
aginaries,  we  readily  arrive 
at  the  following  theorem : 
-X  The  modulus  of  the  sum  of 
fwo  or  more  complex  numbers 
is  less  than,  or  at  most  equal  to,  the  sum  of  their  moduli.  In 
other  words,  a  straight  line  joining  two  points  is  shorter  than 
the  sum  of  the  parts  of  a  broken  line  connecting  the  same  two 
points. 

24.    Multiplication  of  Complex  Numbers.     The  product  of 

z  =  a  +  ib  =  p(cos  6  +  i  sin  0) 
and  z'  =  a'  +  ib'  =  p'(cos  #'  +  *  sin  ^0 

may  be  defined  as  follows : 

z  -  z1  =  pp'  \  cos(0  +  6'}  + 1  sin(0  +  0') } , 

that  is,  the  modulus  of  the  product  ofz  and  z'  is  equal  to  the  prod- 
uct of  their  moduli;  the  amplitude  of  their  product  is  equal  to  the 
sum  of  their  amplitudes. 

Ex.  1.  To  what  power  n  must  z  =  p(cos45°  +  i  sin  45°)  be  raised,  in 
order  that  zn  may  have  the  same  direction  as  z  ?  What  are  the  conditions 
that  zn  =  z  ? 

Ex.2.  Prove  De  Moivre's  Theorem:  (cos  6  +  isin  0)m  =  cosmtf 
+  i  sin  m9,  for  the  case  when  m  is  a  positive  integer. 


SOME   ELEMENTARY   PROPERTIES   OF   EQUATIONS      25 

^ 

25.  Continuity  of  /(z).  We  wish  to  prove  that  /(z)  varies 
continuously  with  z,  that  as  the  complex  number  z  changes 
gradually  from  a  +  ib  to  a'  +  ib',f(z)  changes  gradually  from 


Let  z  vary  from  z0  =  a  +  ib  to  z0  +  h,  where  h  is  likewise  a 
complex  number.     The  corresponding  increment  of  /(z)  is 


and  this,  by  Taylor's  Theorem,  §  18,  is  equal  to 
7>2  J)3 


where  /'(z0),  f"(z0"),  •••,  /"(z0)  are  each  finite  complex  numbers. 
Now,  expression  I  is 


Since  each  term  within  the  parenthesis  of  II  is  a  finite  complex 
number,  and  the  number  of  terms  is  also  finite,  it  follows  that 
the  entire  expression  within  the  parenthesis  has  a  finite  value. 
For,  by  §  23,  the  modulus  of  the  sum  of  two  or  more  complex 
numbers  cannot  exceed  the  sum  of  their  moduli,  and  no  complex 
number  with  a  finite  modulus  can  be  infinite,  no  matter  what 
its  amplitude  (direction)  may  be.  Hence,  by  §  24,  as  the 
modulus  of  h  is  allowed  to  approach  the  limit  zero,  the  modu- 
lus of  the  entire  expression  II  approaches  the  limit  zero.  But 
when  the  modulus  approaches  the  limit  zero,  the  complex  varia- 
ble itself  approaches  zero,  no  matter  what  its  amplitude  may 
be.  Hence  the  expression  II  approaches  the  limit  zero  when 
h  does. 

Since  expression  II  represents  the  difference  between  /(z0+/i) 
and  /(z0),  it  follows  that  an  infinitely  small  variation  of  the 
complex  variable  z  corresponds  to  an  infinitely  small  variation 
of  the  polynomial  /(z),  and  the  continuity  of  /(z)  is  established. 


26  THEORY   OF    EQUATIONS 

The  above  reasoning  remains  valid  if  we  write  the  real  varia- 
ble x  in  place  of  the  complex  variable  z.  For,  real  numbers  are 
only  special  cases  of  complex  numbers. 

An  examination  of  the  graphs  in  §  16  shows  that  when  x  in- 
creases, f(x)  does  not  necessarily  increase ;  it  may  increase  or 
decrease.  What  we  have  proved  is  that,  whether  increasing  or 
diminishing,/^)  passes  from  one  value  to  another  continuously, 
never  per  saltum. 

26.  Fundamental  Theorem.  We  shall  now  demonstrate  the 
important  theorem  which  was  assumed  without  proof  in  §  7,  a 
theorem  which  has  been  called  the  fundamental  proposition  of 
algebra.* 

Every  rational  integral  equation  with  real  or  complex  coefficients 
has  at  least  one  root. 

If  we  can  show  that  the  theorem  is  true  for  the  special  case 
in  which  the  coefficients  of  the  given  equation  are  all  real,  then 
the  general  case,  in  which  some  or  all  of  the  coefficients  are 
complex,  easily  follows.  For,  if  /(z)  is  a  function  of  z,  whose 
coefficients  are,  respectively,  the  conjugate  imaginaries  of  the 
coefficients  of  a  second  function  /2(z),  then  we  may  write 
f,(z}=A+iB  and  f2(z)  =  A~iB,  and  /,(«)  •f,(z}=A^+Ez=f(z}, 
where  /(z)  has  only  real  coefficients.  Now,  if  /(z)  =  0  can  be 
shown  to  have  a  root  a^  then  we  'must  have  either  /i(«i)  =  0 
or  /,(«,)  =0.  Suppose  /1(«])=0,  then  it  follows  that  />(«<>)  =  0, 
where  «2  is  the  conjugate  of  alt  §  8.  Hence  /^z)  =  0  and 
f2(z)  =  0  have  each  at  least  one  root. 

Without  loss  of  generality  we  may  now  assume  that  the 

*  For  historical  and  critical  remarks  on  the  numerous  proofs  which  have 
been  given  of  this  theorem,  see  the  Encykloptidie  d.  Math.  Wiss. ,  I B 1  a,  No.  7 ; 
see  also  Moritz  in  Am.  Math.  Monthly,  Vol.  10,  p.  159.  Gauss  gave  four 
proofs  of  this  theorem,  the  fourth  (1849)  being  a  simplification  of  the  first 
(1799).  The  one  given  here  is  in  substance  Gauss's  proof  of  1849.  It  is 
geometrical  in  character,  and  is  open  to  the  objection  raised  in  the  foot-note 
of  §22. 


SOME    ELEMENTARY    PROPERTIES   OP   EQUATIONS       27 

polynomial  f(z)  of  the  nth  degree  has  real  coefficients  only. 
We  wish  to  prove  that  there  exists  always  at  least  one  value 
of  z,  either  real  or  complex,  which  causes  the  polynomial  /(z) 
to  vanish. 

/Let  z  =  x  +  iy,  then,  by  §  22,  the  variable  represents  points  in 
a'  plane,  and  the  function  f(z)  has  a  definite  value  at  each 
.point  in  the  plane.  As  in  §8,  we  may  write  /(z)  =  P  +  iQ, 
(where  P  and  Q  are  functions  of  x  and  y  with  real  coefficients. 
(To  find  expressions  for  P  and  Q,  let  x  =  r  cos  <£,  y  =  r  sin  <£. 
IBy  De  Moivre's  Theorem, 

zm  =  rm(cos  <£  +  i  sin  <£)"•  =  ^(cos  m<£  +  i  sin  m<£). 
Substituting  for  z  in  /(z),  we  get, 

P  =  rn  cos  n<f>  +  ov""1  cos  (n  —  1)  <£  +  a2r"~2  cos  (w  —  2)  <£  +  •  •  •  +  an, 
l  sin  (n  —  1)  <f>  +  a2r"-2  sin  (n  —  2)  <£  H  ---- 


A  second  expression  for  P  and  Q  is  obtained  by  letting 
t  =  tan  ^  <£.     We  obtain, 


This  gives, 

)  =  r"(l  -f  ft)2" 


If  we  expand  the  binomials  by  the  binomial  formula,  and 
arrange  the  result  according  to  the  powers  of  t,  we  get, 


"(l  +  i2)"'        "(l  +  f2)"' 

where  g'(^)  and  7i(i)  are  rational  integral  functions  of  t,  the 
degrees  of  which  do  not  exceed  2  n. 

All  points  in  the  plane  having  the  same  value  for  r  lie  upon 
a  circle  of  radius  r,  the  centre  of  which  is  at  the  origin  of 


28 


THEORY    OF    EQUATIONS 


coordinates.  To  determine  the  points  on  this  circle  for  which 
P  and  Q  vanish,  we  must  solve  the  equations  </(Z)=0  and 
/i(£)  =  0,  for  the  given  value  of  r.  But  we  know  by  §  7  that  if 
h  (£)  =  0  and  g  (f)  =  0  have  roots  at  all,  they  cannot  have  more 
than  2n.  From  this  it  follows  that  neither  P  nor  Q  can  be 
equal  to  zero  at  all  points  of  an  area  in  the  plane,  for  in  that 
event  we  could  select  r  such  that  the  circle  would  pass  through 
that  area,  and  P  and  Q  would  vanish  at  an  infinite  number  of 
points  on  this  circle. 

The  value  of  Q  may  be  written, 

Q  =  r*/sin  n<f>  +  ^  sin  (n  - 1)  «#>  +  ^  sin (n  -  2)  <£  +  ...\ 

From  this  expression  it  is  readily  seen  that  r  may  be  taken 
so  large  that  Q  has  the  same  sign  as  sin  n<j>  on  all  points  of  the 
circle  where  sin  rt</>  is  numerically  larger  than  some  value  e, 
which  may  be  as  small  as  we  please,  but  not  zero.  Mark  on 
the  circle  the  points 


and  designate  them,  respectively,  by  0,  1,  2,  •••,  2n— 1.     Thus, 
the  circle  is  divided  into  2  n  arcs,  (01),  (12),  (23),  — ,  (2  n— 1,  0), 

in  which  sin  ?«/>  is  alternately 
+  and  — .  The  figure  shows 
the  division  for  n  =  5.  In 
passing  from  arc  (01)  to  arc 
(12),  the  function  Q,  for  suf- 
ficiently large  values  of  r, 
changes  from  +  to  — .  Since 
by  §  25,  Q  is  a  continuous 
function  having  real  values, 
in  going  along  the  circle  from 
+  to  — ,  it  must  at  the  point 
1  pass  through  zero.  Simi- 
larly, Q  must  pass  through 


SOME   ELEMENTARY   PROPERTIES    OF    EQUATIONS       29 

zero  also  at  the  points  2,  3,  •  •  •,  (2  n  —  1),  but  it  does  this  at  no 
other  points  of  the  circle. 

Similar  remarks  apply  to  P.  It  is  readily  seen  that,  for 
sufficiently/large  values  of  r,  P  and  cos  n<j>  have  always  equal 
signs;  that  P  is  positive  at  the  points  0,  2,  •  •-,  (2 n— 2),  and  in 
their  vicinity,  and  negative  at  the  points  1,  3,  5,  •••,  (2  71  —  1), 
and  in  their  vicinity. 

We  have  seen  that  Q  cannot  vanish  at  all  points  of  an  area. 
Consequently  the  area  within  the  circle  can  be  divided  into 
districts  so  that  in  some  districts  Q  is  everywhere  positive, 
while  in  others  it  is  everywhere  negative.  These  districts  are 
marked  on  by  boundary  lines  along  which  Q  vanishes.  To 
aid  the  eye,  "the  positive  districts  are  shaded. 

An  arc  (2 h,  2h  + 1)  of  the  circle,  along  which  Q  is  positive, 
lies  in  a  positive  district.  This  district  lies  partly  inside  and 
partly  outside  the  circle.  Designate  by  /  the  part  of  it  that 
is  inside.  Several  cases  may  arise.  The  area  /may  terminate 
inside,  as  does  (2, 12,  3),  in  which  case  (2  h,  2h  +  1}  is  the  only 
arc  of  the  circle  on  its  boundary.  Or,  the  area  I  may  run  into 
another  positive  arc  (2k,  2&  +  1),  or  it  may  divide  into  two 
or  more  branches,  each  of  which  terminates  in  a  positive  arc 
(21,  2 1  + 1).  If  there  could  be  within  7  an  area,  like  an 
island,  in  which  Q  were  negative,  then  the  conclusions  which 
we  are  about  to  draw  would  still  follow. 

Consider  the  boundary  line  within  the  circle,  passing  from 
2  h  4- 1  to  2k.  Along  this  line  Q  =  0.  But  P  is  negative  at 
the  point  2  h  + 1  and  positive  at  the  point  2  k.  Since  P  is 
continuous  and  represents  real  values,  P  must  pass  through 
zero  in  at  least  one  point  along  the  boundary  line  connecting 
2  h  + 1  and  2  ft.  Thus,  at  that  point,  we  have  not  only  Q  =  0 
but  also  P  =  0 ;  that  is,  f(z)  =  P+iQ  =  Q.  Thus  the  existence 
of  at  least  one  root  of  f(z)  =  0  is  demonstrated. 

The  figure  on  the  preceding  page  is  taken  from  H.  Weber 
and  represents  approximately  the  relations  for  the  equation 


30  THEORY    OF   EQUATIONS 

Its  roots  are  approximately 
«  =  1.52,  p  =  -  .51,  y  =  -  1.24,  €  =  .12  +  i  1.44,  t  =  .12  -  i  1.44 

The  root  «  lies  on  the  boundary  (1,  10,  0). 
The  root  /3  lies  on  the  boundary  (9,  10,  11,  6). 
The  root  y  lies  on  the  boundary  (5,  11,  4). 
The  root  c  lies  on  the  boundary  (3,  12,  2). 
The  root  e'  lies  on  the  boundary  (7,  13,  8). 


CHAPTER   II 

ELEMENTARY  TRANSFORMATIONS  OF  EQUATIONS 

27.  Frequently  it  becomes  necessary  to  transform  a  given 
equation  into  a  new  one  whose  roots  (or  coefficients)  bear  a 
given   relation  to  the  roots  (or  coefficients)   of   the  original 
equation.     The  discussion  of  the  properties  of  an  equation  is 
often  facilitated  by  such  transformations. 

28.  Change  of  Signs  of  Roots.     To  change  an  equation  into 
another  whose  roots  are  numerically  the  same  as  those  of  the 
given  equation,  but  opposite  in  sign,  it  is  only  necessary  to 
substitute  in  the  given  equation  —  x  for  x.     This  transformation 
has  been  used  already  in  the  application  of  Descartes'  Rule 
of  Signs  to  negative  roots,  §  12.     The  signs  of  all  the  terms 
containing  odd  powers  of  x  are  changed  by  it.     The  proof  is 
as  follows  : 

Let  a  be  any  root  of  the  equation  f(x)  =  0.  Then  we  must 
have  /(«)  =  0.  If,  now,  we  substitute  —  x  for  x,  we  get 
/(—  x)  =  0.  Of  this  equation  —  a  is  a  root,  for  when  we  take 
x  =  —  a,  we  have  /(—  [—  «])  =/(«),  and  this  we  know  to  be 
equal  to  zero. 

29.  Roots  multiplied  by  a  Given  Number.     To  transform  an 
equation  into  another  whose  roots  are  m  times  that  of  the  first. 

Put  y  =  mx,  and  substitute  —  for  x  in  the  identity 

m 

a^n  +  a^"-1  +  ...  +  a,  =  a0(x  -  a^(x  —  «2)  —  (*  —  «»)  =  0  ; 
we  get 


mn         mn~'  m  m 

31 


32  THEORY    OF   EQUATIONS 

Multiplying  by  m",  we  have 

Ooyn  +  ma^'1  +  •••  +  mnan^a0(y~ma1~)(y—ma^)  •  •-  (y—ma^)  =0, 
which  is  the  required  equation. 

Hence,  multiply  the  second  term  by  m,  the  third  by  m2,  and 
so  on. 

Ex.  1.    Transform  the  equation  x3  +  ^x2  +  ^x  +  £  =  0  into  an  equa- 
tion with  integral  coefficients  and  a0=  1. 

Multiply  the  roots  by  m  and  we  get  x8  +  —  x2  +  —  x  +  —  =  0.     The 
fractions  will  disappear  if  we  take  m  =  6.     The  result  is 
zs  +  3  X2  +  12  x  +  54  =  0. 

Ex.  2.    Find  the  equation  whose  roots  are  5  times  the  roots  of  the 
equation  x4  —  x3  +  x2  —  x  +  \  =  0. 

Ex.  3.   Find  the  equation  whose  roots  are  —  \  times  the  roots  of 
x4  +  4x3  -  4x2  +  8x  +  32  =  0. 

Ex.  4.    Transform  the  equation  3  x3  +  4  x2  —  5  x  +  6  =  0  into  one  in 
which  the  coefficient  of  x8  is  unity  and  all  coefficients  are  integral. 

Divide  the  left  member  of  the  given  equation  by  3,  then  multiply  the 


roots  by  m.     We  obtain  3?  +        x2  -  —  x  +        -  =  0. 

o  3  3 

Taking  m  =  3,  we  get  the  required  equation,  x3  +  4x2  —  15  x  +  54  =  0. 

Ex.  5.    Change  the  signs  of  the  roots  of  the  equation 

x6  +  5x3-6x2  +  x  +  5  =  0. 
Ex.  6.    Remove  the  fractional  coefficients  from  the  equation 

*+i«r.tW*^ 

keeping  a0  =  1 . 

Ex.  7.  Transform  the  equation  10  x*  —  6  xz  +  1  x  —  ^  =  0  so  that  the 
coefficient  of  the  highest  term  is  unity. 

Ex.  8.  Remove  fractional  coefficients  from  |  x*  -f  \  x8  —  x  +  £  =  0, 
also  make  the  coefficient  of  the  highest  term  unity,  and  change  the  signs 
of  the  roots. 


ELEMENTARY   TRANSFORMATIONS    OF    EQUATIONS      38 

30.  Reciprocal  Roots.  To  change  an  equation  into  a  new  one 
whose  roots  are  the  reciprocals  of  the  roots  of  the  first  equa- 
tion. In  the  equation 

aw?  +  OiX"-1  +  •••  +  an  =  a0(x  —  a^(x  —  «2)  •••  (x  —  «„)  =  0 
put  x  =  -,  and  we  have 

y  ^/ 

••  +a  =  aJ-—         -  —  «     ...-_  «=0. 


y 

J 

Multiplying  by  yn, 

-a^/n  +  an_12/"-1  H  -----  1-  «o  =  a0an(  y  --  }{y  --  )  •  •  •  (  y  -  -  }  =  0, 

V       «i/V        a*J      \       aJ 
the  required  equation. 

31.  Reciprocal  Equations.  If  an  equation  is  not  altered  when 
x  is  changed  into  its  reciprocal,  it  is  called  a  reciprocal  equation. 
Comparing  coefficients  of  the  first  and  last  equation  in  §  30,  we 
see  that  the  conditions  for  a  reciprocal  equation  are 


a0 

The  last  condition  gives  a,t2  =  a02  and  an  =  ±a0.  If  an  =  +«o, 
then  the  denominators  in  the  equations  of  condition  are  all 
alike,  and  we  see  that  the  first,  second,  third  coefficients,  etc., 
taken  from  the  beginning,  are  equal  respectively  to  the  first,  second, 
and  third  coefficients,  etc.,  taken  from  the  end.  If  an  =  —  a0,  then 
these  relations  are  modified  in  this,  that  corresponding  terms 
from  the  beginning  and  end  have  opposite  signs. 

If  a  is  a  root  of  a  reciprocal  equation,  -  must  be  a  root  also. 

«  1 

Hence  the  roots  of  a  reciprocal  equation  occur  in  pairs  a^  —  ; 

I  «i 

«2,  —  ;  etc. 

«2 

If  the  degree  of  the  equation  is  odd,  then  one  of  the  roots 

D 


34  /  THEORY    OF    EQUATIONS 


must  be  its  own  reciprocal  ;  that  is,  one  of  the  roots  must  be 
either  +  1  or  —  1.  If  the  coefficients  have  all  like  signs,  then 
—  1  is  a  root  ;  if  the  coefficients  of  the  terms  equidistant  from 
the  first  and  last  have  opposite  signs,  then  +  1  is  a  root.  In 
either  case  the  degree  of  the  equation  can  be  depressed  by 
unity,  if  we  divide  f(x)  by  x  +  1  or  by  x  —  1.  TJie  depressed 
equation  is  always  a  reciprocal  equation  of  even  degree  Kith  like 
signs  for  its  coefficients. 

If  the  degree  of  a  given  reciprocal  equation  is  even  and  if 
terms  equidistant  from  the  first  and  last  have  opposite  signs, 
then  the  left  member  of  the  equation  has  xr  —  1  as  a  factor. 
For,  the  equation  may  be  written  in  the  form 

(^»  -  1)  +  alx(x2n-2  -  1)  +  a2"-4  -  1)  +  t"  =  0. 


Dividing  by  x2  —  1  reduces  this  type  of  reciprocal  equation  to 
one  of  even  degree  with  all  coefficients  positive. 

Since  all  reciprocal  equations  of  odd  degree  and  all  recipro- 
cal equations  of  even  degree  with  half  of  the  coefficients 
negative,  are  reducible  to  reciprocal  equations  of  even  degree 
with  coefficients  ail  positive,  the  latter  kind  is  called  the 
standard  form  of  reciprocal  equation. 

Ex.  1.     Under  what  conditions  is  the  equation 

**  +  aix3  +  a2x2  +  asx  +  a4  =  0  reciprocal  ? 
Under  what  conditions  is  it  in  the  standard  form? 
Ex.  2.     Reduce  the  following  reciprocal  equation  to  the  standard  form. 
x6  +  «ix5  4-  a2x*  —  atf?  —  a\x  —1=0. 

We  may  write  it  thus  :  (x6  -  1)  +  aix(x*  -  1)  +  a2x2(x2  -  1)  =  0. 
Dividing  by  a;2  -  1,  x4  +  aix3  +  (1  -f  a2)  x2  +  a\x  +  1  =  0. 

Ex.  3.     For  what  value  of  am  will 

x2m  +  aix2™-1  +  a2x2n»-2  +  •••  +  amxm  -  am.ixm~l  -  •••  -  a&  -  1  =  0 
be  a  reciprocal  equation  ? 

Ex.  4.     Solve  the  equation  x4  +  3xs-3x-l=0. 


ELEMENTARY   TRANSFORMATIONS    OF   EQUATIONS      35 


Ex.  S.    Solve  the  equation  3x3  +  2x2 
Ex.  6.     Given  that  c  is  a  root  of 

ax5  +  (b  —  ac)  x4  —  bcxs  —  bx2  —  (a  —  be)  x  +  ac  =  0, 
find  the  other  roots. 

32.  Roots  diminished  by  a  Given  Number.  If  an  equation  is 
to  be  transformed  into  another  whose  roots  are  those  of  the 
first,  diminished  by  h,  then  we  take  y  =  x  —  h,  and  substitute 
x  =  y  -f-  h  in  the  given  equation 


a<pn  +  a^"-1  H  -----  1-  «„  =  0.  I 

We  obtain     a0  (y  +  li)n  +  «i  0  +  /i)""1  +  —  +  an  =  0.  II 

If  a  is  a  root  of  equation  I,  then  a  —  h  is  a  root  of  equation  II  ; 
for,  substituting  a  —  h  for  y  in  the  latter,  we  get 

a0an  +  a^an~l  +  •••  +  an, 

which  expression  must  vanish,  since  a  is  a  root  of  I.     Hence 
II  is  satisfied  by  y  =  a  —  h. 

If  we  expand  the  binomials  in  II  and  collect  the  coefficients 
of  like  powers  of  y,  we  obtain,  let  us  suppose,  the  equation 


A#f  +  An/*-1  +  A%»-2  +  -  +  A  =  0. 
Since  y  =  x  —  h,  this  equation  is  equivalent  to 

iO  -  K)  +  An  =  0. 


The  form  of  this  last  equation  suggests  an  easy  rule  for  carry- 
ing out  the  actual  computation.  Dividing  the  left  member  by 
x  —  h,  the  remainder  obtained  is  seen  to  be  equal  to  An,  the 
absolute  term.  If  the  quotient  thus  obtained  is  divided  by 
x  —  h,  the  remainder  is  An_^  the  coefficient  of  x.  By  continu- 
ing this  process  we  can  find  all  the  coefficients  of  the  trans- 
formed equation. 

If,  instead  of  diminishing  the  roots,  we  desire  to  increase 
them,  we  take  h  negative. 


36  THEORY   OF   EQUATIONS 

Ex.  1.  Transform  x4  —  5z3-f7z2  —  4x  +  5  =  0  into  another  equation 
whose  roots  are  less  by  2. 

By  synthetic  division  the  process  is  as  follows  : 

1     -5  +7  _4     4-5  LJ 

+2  -6  +2     -4 

-3  +1  -2     +1 

+2  -2  -2 

-1  -1  -4 

+  2  +2 

+  1  +1 

±2 

+  3 

The  numbers  in  black  type,  1,  —  4,  -f  1,  +3,  indicate,  respectively,  the 
first,  second,  third,  and  fourth  remainder.  Hence  the  required  equation 
is  x*  +  3  xs  +  z-  -  4  x  +  1  =  0. 

Ex.  2.    Diminish  the  roots  of  2  x5  -  Xs  +  10  x  -  8  =  0  by  5. 

Ex.  3.  Transform  the  equation  x4  —  8  x3  +  a:2  +  x  —  6  =  0  into  another 
in  which  the  second  term  is  wanting. 

The  sum  of  the  roots  of  the  given  equation,  by  §  13,  is  +  8.  In  the 
required  equation  the  sum  shall  be  zero.  Hence  the  sum  of  the  roots 
must  be  diminished  by  8  ;  each  single  root  by  2.  Hence  we  get  by 
synthetic  division  x*  -  23  x2  -  59  z  -  48  =  0. 

Ex.  4.    Remove  the  second  term  of  x5  +  10  x4  +  x2  +  1  =  0. 
Ex.  5.    Remove  the  second  term  of  4  x4  +  8  x3  +  x  +  12  =  0. 

33.  Removal  of  Second  Term  in  the  Cubic.  In  the  transfor- 
mation of  the  general  cubic 

=  0 


into  another,  deprived  of  the  second  term,  we  notice  that  each 
root  must  be  increased  by  ^,  the  sum  of  the  roots  in  the  given 

O    1.  ^0  /!  J) 

cubic  being  --  j—  ?•     Put  y  =  x  -f  y^,  then  x  =  y  —  ^-    Substitut- 

00  GO  00 

ing,  we  obtain 

+  3  b*y  - 


ELEMENTARY    TRANSFORMATIONS   OF    EQUATIONS      oT 

ExpafMing,   and   collecting   the   coefficients   of   the   different 
powers  of  y,  we  get 


0, 
where  b0R2  =  b0b2  —  6t2  =  H, 

V^s  =  V&s  -  3  b0b,b.2  +  2b,s  =  G. 

Accordingly,  the  transformed  cubic,  deprived  of  the  second 
term,  is 


If  the  roots  of  this  equation  are  multiplied  by  &0,  by  the 
process  shown  in  §  29,  and  the  letters  H  and  G,  as  defined 
above,  are  introduced  for  brevity,  then  the  transformed  cubic 
takes  the  form 


Since  z  =  b0y  and  y  =  x-\-^,  we  have  z  =  b<fc  +  bj. 
fy> 

The  reader  will  observe  that  by  the  use  of  the  binomial 
coefficients,  1,  3,  3,  1,  in  the  original  cubic,  the  expressions 
arising  in  the  process  of  transformation  are  simplified  some- 
what. The  use  of  binomial  coefficients  is  frequently  found 
convenient. 

34.  Removal  of  Second  Term  in  the  Quartic.  Write  the 
quartic  with  binomial  coefficients,  thus, 

b<p*  +  4  &!»>  +  6  62**  +  4  bsx  +  64  =  0. 

The  sum  of  the  roots  being  —  j-1,  each  root  must  be  increased 

by-1-     Putting  y  =  x  +  r1,  we  have  x  =  y  —  ^~     Substituting 
o0  60  o0 

in  the  quartic  and  expanding  the  binomials,  we  obtain 

s 


4-  6  bnbfb»  —  3  6,4)  =  0, 

7      A  \~\l    ~*  V"*"O        1^  VU^1    V2  1    / 

^0 


38  THEORY    OP    EQUATIONS 

where  H  and  G  are  defined  in  §  33.  The  last  term  of  the 
transformed  quartic  it  is  most  convenient  to  consider  as 
composed  of  H  and  of  a  new  function  /.  Let  /=  b0bt  — 
4  &A  +  3  622-  Then  we  obtain  the  following  : 

603&4  -  4  V&i&s  +  6  &A2&2  -  3  V  =  V  (&A  -  4  6A  +  3  b/) 

-  3  (b0b.2  -  6f)2  =  V/-  3  H2. 
The  transformed  quartic  takes  now  the  form 

,    6   rr  2  ,    4  ~     .  b02I-3H2     n 
f+-Htf+-Gy  +  -  —  --  =  0,  I 

00  C/o  t>0 

or,  multiplying  the  roots  by  60,  the  form 


Since  z  =  &0y,  and  y  =  x  +  ~)  we  have  2  =  b<fe  +  6^ 

c*o 

Ex.  1.  Compute  H  and  £  for  the  cubic,  obtained  by  transforming 
Xs  +  3  x2  +  4  x  —  10  =  0,  so  that  the  second  term  will  vanish. 

Ex.  2.  Compute  H,  G,  and  7  for  the  quartic  with  the  second  term 
wanting,  obtained  from  2  x4  —  16  x3  —  2  x2  -f  x  —  12  =  0. 

Ex.  3.  Verify  the  results  obtained  in  the  last  two  exercises  by  trans- 
forming the  cubic  and  quartic  by  the  process  of  synthetic  division,  as 
in  §  32. 

35.   Equation  of  Squared  Differences  of  Roots  of  Cubic.     The 

formation  of  the  equation  whose  roots  are  the  squares  of  the 
differences  of  every  two  of  the  roots  of  a  given  cubic  is  of 
importance,  because  the  equation  thus  formed  leads  with  com- 
parative ease  to  the  criteria  of  the  nature  of  the  roots  of  the 
general  cubic.  Let  the  cubic  be 

fcoor5  +  3  b^  +  3  b^x  +  63  =  0.  I 

Transforming  so  as  to  remove  the  second  term,  we  have,  by  §  33, 


where 


ELEMENTARY   TRANSFORMATIONS   OF   EQUATIONS      39 

Let  the  roots  of  equation  II  be  «,  ft,  y.  Then  the  squares  of 
the  differences  of  every  two  of  the  roots  are 

(a -ft)2,    («-y)2,    (ft-y?.  Ill 

Since  the  roots  of  II  are  the  roots  of  I,  each  increased  by  — ,  it 

b0 

follows  that  the  differences  of  the  roots,  two  by  two,  of  equa- 
tion II  are  the  same  as  the  differences  of  the  roots  of  equation  I. 
Hence  the  squares  of  the  differences,  given  in  III,  are  the 
squares  of  the  differences  of  the  roots  of  equation  I,  as  well  as 
of  equation  II.  In  other  words,  both  equations  lead  to  the 
same  "  equation  of  squared  differences."  This  last  equation  is 
evidently 

Jz-(«-/Wz-(«-y)2Hz-03-y)2f  =  °-        iv 

The  coefficients  may  be  calculated  as  follows :  Equation  IV  is 
satisfied  by  the  equality 

z  =  (a  -  ft)2.  -  X  ' 
We  obtain  from  this 


Now  a2+/82+y2  was  shown  in  §  15,  Ex.  2,  to  be  equal  to  al2—2a2 : 

O     TT 

in  the  case  of  equation  II,  a^  =  0,  a2  =  —j-     So, 

f)n 


-" 


while  afty  =  —  —  • 

n  ° 
°0 

Hence  we  may  write 

6H      22 


where  y2  and  y  are  written  for  y2  and  y.  This  is  allowable, 
since  y  is  one  of  the  three  possible  values  that  y  can  assume  in 
equation  II. 


40  THEORY    OF    EQUATIONS 


Multiplying  the  members  of  the  last  equation  by  y,  we  have 


Subtracting  equation  II  from  this,  we  get 

3H       3G     A 

—  -y-  —  =  0, 

"0  yO 

36? 


whence  y  = 

We  have  here  y  expressed  as  a  linear  function  of  z.  Sub- 
stituting this  expression  of  y  in  equation  II,  we  obtain,  after 
some  labor, 

-\  O    TT  Q-l      TT2  0*7 

V 


This  is  the  "  equation  of  squared  differences  "  of  the  roots  of 
equation  I  and  of  equation  II,  the  roots  of  V  being 

(a-/3)2,    («-y)2,    (/?  -y)2. 

Multiplying  the  roots  of  equation  V  by  602,  we  obtain  an  equa- 
tion free  of  fractions, 

23  +  18  Hz-  +  81  H2z  +  27  (G2  +  4  fi"3)  =  0,  VI 

whose  roots  are 

&02(«-/?)2,  &02(«-r)2,  W-y)2. 

Here     (a  -  /8)2(a  -  y)208  -  y)2  =  -  |I  (G2  +  4  H*)  =  D, 

where  D  is  an  important  function,  known  as  the  discriminant 
of  the  cubic.     Since,  by  §  33, 


we  obtain 

bJD  =  27  (3  W  +  6  &0&A&3  -  V&32  -  4  W  -  4  6^6,). 

In  the  discussion  of  the  cubic  equation  we  shall  frequently 
make  use  of  the  discriminant. 


ELEMENTARY   TRANSFORMATIONS    OF   EQUATIONS      43 

Ex.  1.  Find  the  equation  of  squared  differences  of  the  roots  of  the 
cubic  xs  +  3  x2  -  3  x  -  1  =  0. 

Here  60  =  1,  61  =  1,  62  =  -  1,  63  =-  1.  Hence  G-  =  4  and  H=-2, 
The  required  equation  is  z3  -  36  z*  +  324  g  -  432  =  0. 

Ex.  2.  The  cubic  in  the  previous  example  is  a  reciprocal  equation. 
Solve  it,  find  the  values  of  the  squared  differences  of  the  roots,  and  see 
whether  they  are  really  roots  of  the  equation  of  squared  differences. 

The  reciprocal  equation  of  the  standard  form,  obtained  from  the  above, 
is  a;2  +  4  x  +  1  =  0.  The  roots  of  the  given  cubic  are  1,  —  2  ±  V3  ;  their 
squared  differences  are  12,  12  ±  6\/3.  Dividing  the  left  member  of  the 
transformed  cubic  by  z  —  12,  thus, 

1  _  36  +  324  -  432  Q2 

+  12  -  288  +  432 
_24+36+      0 

we  see,  by  §4,  that  12  is  a  root.     The  depressed  equation,  z"1— 24  2+36=0, 
is  satisfied  by  z  =  12  ±  6  V3. 

Ex.  3.  Find  the  equation  of  squared  differences  of  the  roots  of  the 
cubic  Xs  +  x2  -  x  -  1  =  0. 

The  required  equation  is  z3  —  8  22  +  16  z  —  0.  What  inference  can  be 
drawn  with  respect  to  the  roots  of  the  given  cubic  from  the  fact  that  z  =  0 
is  a  root  of  the  transformed  cubic  ?  ^  -y.  rO'-rf  3  />/  c(.(fat<^ 

Ex.  4.  Find  the  equation  of  the  squared  differences  of  the  roots  of 
*8  +  3  x  +  2  =  0.  Am.  z*  +  18  z2  +  81  z  +  216  =  0. 

It  is  important  to  observe  that,  since  the  last  term  +  216  is  positive, 
and  is  equal  to  minus  the  product  of  the  roots,  at  least  one  of  the  three 
values  of  z  must  be  negative.  Now  if  the  roots  of  the  given  cubic  are  all 
real,  then  the  squares  of  their  differences  must  be  positive,  and  all  the 
values  of  z  must  be  positive.  A  negative  value  of  z  can  be  obtained  only 
when  the  given  cubic  has  two'imaginary  roots.  Hence  x3  +  3o;  +  2  =  0 
has  two  imaginary  roots.  Verify  this  by  Descartes'  Rule  of  Signs. 

Ex.  5.  Find  the  equation  of  the  squared  differences  of  the  roots  of 
Xs  +  6  x2  +  5  x  -  16  =  0. 

The  process  is  easier  if  we  first  transform  the  cubic  to  another  whose 
second  term  is  wanting. 

36.  Criteria  of  the  Nature  of  the  Roots  of  the  Cubic.  We  pro- 
ceed to  discuss  the  nature  of  the  roots  of  the  general  cubic  I  in 
§  35,  with  the  help  of  the  "  equation  of  squared  differences  "  V. 


42  THEORY    OP   EQUATIONS 

To  begin  with,  observe  that,  since  the  absolute  term  in  V  is 
equal  to  minus  the  product  of  the  three  roots  of  V,  at  least  one 
of  the  three  roots  must  be  negative  when  the  absolute  term  is 
positive.  But  a  negative  root  cannot  occur  in  V,  if  all  the  roots 
in  I  are  real.  A  negative  result  can  be  obtained  only  when  the 
number  that  is  being  squared  is  imaginary.  Hence,  a  negative 
root  in  V  indicates  the  presence  of  two  imaginary  roots  in  I. 

Again,  when  all  the  roots  in  V  are  positive,  then  I  cannot  have 
imaginary  roots.  For,  the  square  of  the  difference  of  two  con- 
jugate imaginary  roots  is  always  real  and  negative,  making  the 
absolute  term  in  V  positive  and  one  of  its  roots  negative. 

Real  Roots.  Equation  I  has  real  roots  when  G~  +  kHz  is 
negative.  For,  to  make  this  negative,  H  must  be  negative  and 
4  H3  must  be  numerically  greater  than  6r2.  That  being  the 

case,  the  signs  of  the  coefficients  in  V  are  H 1 .  Hence, 

by  Descartes'  Rule  of  Signs,  V  can  have  no  negative  roots 
Since  all  these  roots  are  real,  they  must  be  positive.  Conse- 
quently, equation  I  has  all  its  roots  real. 

Complex  Roots.  Equation  I  has  two  complex  roots  when 
Cr2  +  4  H3  is  positive.  For,  when  this  is  positive,  one  of  the 
roots  in  V  is  negative. 

Two  Equal  Roots.  Equation  I  has  two  equal  roots  when 
G2  +  4 H'A  =  0.  For,  in  this  case,  z  =  0  is  a  root  of  V,  showing 
that  two  of  the  roots  in  I  have  zero  for  their  difference.  Thus, 
the  vanishing  of  the  discriminant  indicates  equal  roots. 

TJiree  Equal  Roots.  Equation  I  has  three  equal  roots  when 
H=0  and  G  =  0.  For,  V  reduces  to  z3  =  0.  Since  all  the  roots 
of  V  are  zero,  all  the  roots  of  I  must  be  equal  to  one  another. 

Ex.  1.  Prove  that  equation  V  in  §  35  cannot  have  three  equal  roots 
different  from  zero. 

/- —  Ex.  2.   If  two  roots  in  V  are  equal  to  each  other,  but  not  zero,  what 
inference  can  be  drawn  about  the  roots  of  I  ? 

Ex.  3.    Compute  the  discriminant  of  xs  —  6  a;2  +  3  x  —  4  =  0. 

Ex.4.  Find  the  discriminant  of  4  x3  +  8  x2  +  5x  +  1  =  0.  What 
inference  can  be  drawn  from  its  value  ? 


CHAPTER   III 
LOCATION  OF  THE  ROOTS  OF  AN  EQUATION 

37.  In  this  chapter  we  shall  deduce  theorems  giving  limits 
between  which  all  the  real  roots  of  an  equation  with  real  coeffi- 
cients lie.     We  shall  also  derive  theorems  which  enable  us  to 
separate  from  each  other  all  the  distinct  real  roots,  and  to 
ascertain  the  exact  number  and  location  of  the  real  roots. 

38.  An  Upper  Limit.     If  in  the  equation  f(x)  =  0  the  coefficient 
of  xn  is  unity,  then  the  numerically  greatest  negative  coefficient, 
increased  by  one,  is  an  upper  limit  of  the  positive  roots  of  tt~ 
equation. 

Any  positive  value  of  x  makes  f(x)  >  0,  if  it  makes 
a.«_^a.»-i  +  a,»-2+  ...  +i)>o, 

(Kn  —  1 

or,  x*  —  p  •  -      —  >  0, 

x  —  L 

where  p  is  the  numerical  value  of  the  greatest  negative  coeffi- 
cient.    All  the  more  is  f(x)  >  0,  if  a  positive  value  of  x  makes 


1    x-1 
or,  U"-l)(  l--2-r}>0. 


But  this  last  expression  is  always  ^  0,  or  positive,  if  p  <  x  —  1; 
that  is,  if  x  >  p  + 1. 

Since  any  real  value  of  x,  greater  than  p+l,  makes  /(#)>0, 
every  real  value  of  x  which  makes  f(x)  equal  to  zero  must  be 
equal  to  or  less  than  p  +  1.  Hence  p  + 1  is  an  upper  limit  of 
the  real  positive  roots  of  f(x)  =  0. 

43 


44  THEORY   OP   EQUATIONS 

39.  Another  Upper  Limit.  If  the  numerical  value  of  each 
negative  coefficient  is  divided  by  the  sum  of  all  the  positive  coeffi- 
cients which  precede  it,  the  greatest  of  the  fractions  thus  formed, 
increased  by  one,  is  an  upper  limit  of  the  positive  roots  off(x)  =  0. 


Let  f(x)  = 
in  which  the  coefficients  of  x"~2  and  x"~*  are  negative.     Since 

(xm  -  1)  =  (x  -  1)  (of-1  +  xm~-  +  ...  +  a;  +  1), 
we  have     of  =  (x  -  1)  (af-1  +  xm~2  -\  -----  1-  a;  +  !)  +  !. 

If  we  transform  all  the  positive  terms  in  f(x)  by  means  of 
this  formula,  we  obtain  f(x)  = 


\-al 


If  in  this  expression  x  is  assigned  a  positive  value  large 
enough  to  make  the  sum  of  the  coefficients  in  each  column 
of  terms  positive,  then  f(x)  will  be  positive  for  that  value 
of  x.  The  coefficients  in  the  first  and  third  column  are  posi- 
tive, if  x  >  1.  The  same  is  true  of  all  other  columns  which 
are  free  of  negative  coefficients. 

The  sum  of  the  coefficients  in  the  second  column,  contain- 
ing the  negative  coefficient  —  q2,  is  positive  if  x  is  large  enough 
to  make  '  o^-l)  +o1(a;-l)  -a2>0. 

Whence  x  >  —  ^—  +  1. 


.  Similarly,  we  obtain  from  the  fourth  column,  if 
a0(x  —  1)  +  a,  (a?  —  1)  +  a3(x  —  1)  —  a4  >  0, 


the  inequality  x  >  -  —  —  +  1. 

J  ^^ 


LOCATION   OF   THE   ROOTS    OF   AN    EQUATION          45 

The  same  reasoning  applies  to  any  column  containing  a 
negative  coefficient.  Hence,  if  we  take  x  equal  to,  or  greater 
than,  the  greatest  of  the  expressions  thus  obtained,  then  the 
polynomial  f(x)  will  be  positive,  and  the  greatest  expression 
constitutes  an  upper  limit  of  the  positive  roots. 

Ex.  1.    Find  upper  limits  of  the  positive  roots  of 

x*  -  8  xs  +  18  a;2  -  16  x  +  5  =  0. 
By  §  38,  17  is  an  upper  limit. 

Q  1  /> 

By  §  39,  the  fractional  expressions  are  --f  1   and 1- 1. 

1  1  +  18 

Hence  9  is  an  upper  limit.  The  largest  positive  root  is  5.  Thus  §  39 
gives  here'a  closer  limit  than  §  38.  The  limit  obtained  from  §  38  is  never 
smaller  than  that  obtained  from  §  39,  and  usually  not  so  small. 

Ex.  2.    Find  superior  limits,  by  §  38  and  by  §  39,  of 

(1)  x*  +  45  x2  -  40  x  +  84  =  0. 

(2)  3  x*  +  6  x3  +  12  x2  -  4  x  -  10  =  0. 

(3)  2  x5  +  10  x4  -  72  xs  +  5  x2  +  15  x  -  39  =  0. 

(4)  2  xs  -  5  x2  +  x  +  10  =  0. 

^  40.  Lower  Limits.  A  number  not  greater  than  any  of  the 
positive  roots  of  an  equation  constitutes  a  lower  or  inferior 
limit.  Such  a  limit  may  be  found  by  transforming  the  given 
equation  into  another  whose  roots  are  the  reciprocals  of  the 
roots  of  the  given  equation.  By  §  30,  this  can  be  done  by 

writing  x  =  -•     In  the  transformed  equation  we  find  a  superior 
limit  of  y ;  the  reciprocal  of  y  will  be  an  inferior  limit  of  x. 

41.  Limits  of  Negative  Roots.  Substitute  in  the  given  equa- 
tion —  y  for  x,  and  then  find  the  superior  and  inferior  limits 
of  the  positive  roots  of  the  transformed  equation. 

Ex.  1.  Find  limits  of  the  positive  and  of  the  negative  roots  of 
X*  -  19  x2  -  23  x  -  7  =  0. 

By  §  38  and  §  39  the  upper  limits  are  24.     Writing  -  for  x,  we  get 


46  THEORY   OF   EQUATIONS 

V 

7  y*  +  23  y3  +  19  y2  —  1  =  0.  The  upper  limits  of  the  roots  of  this  equation 
are  f  and  £f ;  hence  the  lower  limits  of  the  positive  roots  of  the  given 
equation  are  |  and  f  g. 

Writing  —  y  for  x,  we  obtain  y*  —  19  y"2  +  23  y  —  7  =  0.  We  obtain  20 
as  a  superior  limit  and  J$  as  an  inferior  limit  of  the  positive  values  of  y. 
Hence  the  negative  roots  of  the  given  equation  lie  between  —  ^  and  —20, 
and  all  the  roots  lie  between  24  and  —20. 

To  convey  an  idea  of  how  the  limits  compare  with  the  actual  values  of 
-a;,  we  give  the  roots:  4.8977  •••,  -3.6331.-.,  -.7124-..,  -.5522-... 

Ex.  2.  Between  what  limits  do  the  real  roots  of  se5  +  5  x*  +  x8  —  16  a;2 
-  20  x  -  16  =  0  lie  ? 

By  §  38  and  §  41,  the  roots  lie  between  21  and  -21.  By  §  39  and  §  41, 
the  roots  lie  between  y  and  —  6.  The  roots  are  2,  —2,  —4,  K-liV— 3). 

Ex.  3.   Between  what  limits  are  the  real  roots  of 

(1)  x*  +  4z3  -  x2  -  16x  -  12  =  0, 

(2)  5C*-3o;3  +  3z-l  =  0, 

(3)  x5  -  11  x4  +  17  xs  +  17  x2  -  11  x  +  1  =  0  ? 

V 

"42.  Change  of  Sign  of  f(x).  If  two  real  numbers  a  and  b,  ivhen 
substituted  for  x  inf(x),  give  to  f(x)  contrary  signs,  an  odd  num- 
ber of  roots  of  the  equation  f(x)  =  0  must  lie  between  a  and  b  ;  if 
they  give  to  f(x)  the  same  sign,  either  no  root  or  an  even  number 
of  roots  must  lie  between  a  and  b. 

Since  f(x)  varies  continuously  with  x  (§  25),  and  f(x)  changes 
sign  in  going  from  /(a)  to  /(&),  passing  through  all  the  inter- 
mediate values,  it  follows  that  f(x)  must  pass  through  the  value 
zero.  That  is,  there  is  some  real  value  of  x,  between  a  and  6, 
which  causes/(o?)  to  vanish  and  is  a  root  of  the  equation /(x) =0. 
But  f(x),  in  passing  from  /(a)  to  /(6),  may  go  through  zero 
nore  than  once.  When  /(a)  and  /(&)  have  opposite  signs,  f(x) 
must  pass  through  zero  an  odd  number  of  times.  Since  a  real 
root  corresponds  to  a  point  where  the  graph  of  f(x)  crosses  the 
axis  of  x,  the  statement  just  made  simply  means  that,  to  pass 
from  a  point  on  one  side  of  the  axis  to  a  point  on  the  other 
side  of  it,  we  must  cross  the  axis  an  odd  number  of  times. 


LOCATION   OP   THE   ROOTS    OF    AN    EQUATION  47 

Similarly,  if  /(a)  and  /(&)  have  like  signs,  they  represent 
two  points  on  the  same  side  of  the  axis.  To  pass  from  one 
point  to  the  other,  the  graph  either  does  not  cross  the  axis  at 
all,  or  it  crosses  the  axis  an  even  number  of  times.  Hence,  if 
f(a)  and  /(&)  have  like  signs,  there  are  either  no  roots  or  an 
even  number  of  roots  between  a  and  6. 

Ex.  1.    Locate  the  roots  of  x4  +  4  z3  —  x"2  —  16  x  —  11  =  0. 
From  Descartes'  Rule  of  Signs  (§  11)  we  see  that  there  cannot  be  more 
than  one  positive  root  and  not  more  than  three  negative  roots.     We  find 


/(I)  =  -28.  /(-2)  = 


We  see  that  the  positive  root  lies  between  1  and  2,  that  the  negative 
roots  lie  respectively  between  0  and  —  1,  —  2  and  —  2.7,  —  2.7  and  —  3. 

Ex.  2.    Locate  the  roots  of  of  —  5  x*  +  9  x3  —  9  x2  +  5  x  —  1  =  0. 
By  Descartes'  Rule  of  Signs  we  see  that  there  are  no  negative  roots. 
We  obtain  6  as  a  superior  limit  of  the  positive  roots.     We  have 


/(.5)  =  +  .09. 

/(I)  =0.  /(6)  =  +2945. 

We  see  that  1  is  a  root  ;  that  there  is  a  root  between  0  and  .5  ;  also 
between  2  and  3.  Two  roots  are  still  unaccounted  for  ;  they  are  imaginary, 
as  can  be  ascertained  by  Sturm's  Theorem,  to  be  given  later. 

Ex.  3.   Locate  the  real  roots  of 

(1)  x3-3x2-46x-71  =  0. 

(2)  x*  +  2  xs  -  41  x2  -  42  x  +  361  =  0. 

(3)  x*  -  16  x3  +  86  x2  -  176  x  +  110  =  0. 

\f  43.  Maximum  and  Minimum  Values  of  f(x).  Any  value  of  x 
which  renders  f(x)  a  maximum  or  a  minimum  is  a  root  of  the 
derived  function  of  f'(x). 

First.  Let  a  be  a  value  which  makes  f(x)  a  minimum. 
Since  /(a)  is  a  minimum,  it  is  less  than  both  f(a  —  h)  and 


48  THEORY    OF   EQUATIONS 

/(a  4-  7i),  where  h  is  a  small  increment.     By  Taylor's  Theorem 
(§  18)  we  have 

f(a  -K)  -/(a)  =-/'(«)•  ft  +  /"(«).  I'—., 


/(a  +  A)  -/(a)  =  +/'(«)•  7i+/"(a)  . 

Since  the  left  members  of  these  equations  are  both  positive, 
the  right  members  must  be  positive  too.  Now  h  may  be  taken 
so  small  that  the  sign  of  the  right  member  of  each  equation  is 
the  same  as  the  sign  of  the  first  term  in  the  right  member. 
Hence  —/'(«•)  •  /*•  and  +/'(«)  -h  must  both  be  of  the  same 
sign.  But  this  is  possible  only  when  /'(a)  =0;  that  is,  when 
a  is  a  root  of  the  first  derivative.  Since  in  each  equation  the 
right  member  is  positive,  and  the  first  term  in  that  member  is 
zero,  it  follows  that  /"(a)  is  positive. 

Second.  Suppose  that  x  =  a  makes  f(x]  a  maximum.  Then 
the  left  members  of  the  above  equations  are  both  negative. 
That  the  right  members  may  be  both  negative,  for  very  small 
values  of  h,  it  is  necessary  not  only  that  /'  (a)  should  vanish  as 
before,  but  that  /"(a)  be  a  negative  value. 

44.  Rule  for  Maxima  and  Minima.  The  proof  of  the  preced- 
ing article  suggests  the  following  rule  for  finding  maximum 
and  minimum  values  of  f(x)  :  Solve  the  equation  f  (x)  =  0. 
Each  of  its  roots  renders  f(x)  a  maximum  or  minimum,  accord- 
ing as  it  makes  f"(x)  negative  or  positive. 

Ex.  1.  Find  the  maxima  and  minima  of  f(x)  =  2xa  +  15  x2  +  36  z+5. 
Here  /'(x)  =  6  x2  +  30  x  +  36, 

and  /"  (x)  =  12  x  +  30. 

~l 
/'  (x)  =  0  gives  x  =  —  2,  or  —  3.     We  find  that  /"  (—  2)  is  positive  and  ;, 

/"(—  3)   is  negative.     Hence  /(—  2)  is  a  minimum   and  /(—  3)  is  a  i 
maximum. 

Ex.  2.  Find  the  maximum  and  minimum  values  of  /(x)  =  2  x3  +  3  xa 
-  36  x  +  75, 


LOCATION   OF   THE   BOOTS   OP   AN   EQUATION 


49 


45.   Rolle's  Theorem.     Between  two  successive  real  roots  a  and. 
6  of  the  equation  f(x)  =  0  there  lies  at  least  one  real  root  of  the 


equation  f  (x)  =  0. 

f*t*S7    CCS,     -     r2oCt/*4 

Let  the  curve  in  this  "figure  be  the  graph  of  f(x)  —  0.     The 

points  A,  B,  C,  D,  E,  F,  G  represent  maximum  and  minimum 
values  of  f(x) ;  the  points  M,  N,  P  represent  real  roots  of  f(x) =0. 
Between  the  two  roots  M  and  JV"the  curve  bends  down  and 


/N 


then  up.  Between  the  real  root  at  JV  and  the  double  root  at 
P  the  curve  goes  up,  down,  up,  and  finally  down.  Evidently, 
between  each  pair  of  distinct  successive  real  roots  there  must 
be  at  least  one  maximum  or  minimum  value  of  /(#)• 

15i it  each  maximum  or  minimum  point  represents  a  value  of 
x  which  is  a  root  of  the  equation /'(o;)j=^0_  (§  44).  Hence 
Rolle's  TheoremTs  proved. 

From  the  examination  of  the  figure  we  see  that  two  successive 
roots  of  the  derived  function  may  not  comprise  between  them 
any  real  root  of  f(x)  =  0,  as  in  case  of  the  roots  represented  by 
D  and  E ;  they  may  comprise  one  distinct  root,  as  in  case  of 
the  roots  at  A  and  B,  B  and  (7,  E  and  F,  but  they  can  never 
comprise  more  than  one  root  of  f(x)  =  0. 

£ 


50  THEORY  OF   EQUATIONS 

Ex.  1.  The  equation  £*  —  12  xs  +  47  x2  —  72  x  +  36  =  0  has  the  root* 
1,  2,  3,  6.  Locate  the  roots  of  the  equation  2  z3  —  18  x2  +  47  x  —  36  =  0 
by  Rolle's  Theorem. 

46.  The  determination  of  the  number  of  real  roots  and  of 
complex  roots  of  an  equation  is  a  problem  which  has  engaged 
the  attention  of  several  great  mathematicians.     Researches  on 
this  subject  have  been  made  by  Descartes,  Newton,  Waring, 
Budan,  Fourier,  Sylvester,  Sturm,  and  some  more  recent  mathe- 
maticians.    Nearly  all  of  the  theorems  and  rules  are  defective  in 
not  giving  the  exact  number  of  real  roots  or  ofjm«ginary  roots, 
but  of  giving  merely  a  superior  liniiff"Eo~this  number.     Des- 
cartes' Rule  of  Signs,  for  instance,  gives  only  superior  limits 
for  the  number  of  positive  and  negative  roots. 

The  theorem  of  Sturm  is  free  from  this  blemish.  It  tells 
always  the  exact  number  of  real  roots  within  a  given  interval 
and  the  exact  number  of  imaginary  roots  of  an  equation.  Be- 
cause of  this  unfailing  certainty  we  select  Sturm's  Theorem  to 
the  exclusion  of  the  theorems  of  Newton,  Sylvester,  Budan,  and 
Fourier,  even  though  it  is  laborious  in  its  application.  In  prac- 
tice, the  nature  and  situation  of  the  roots  are  more  usually  found, 
when  possible,  by  the  theorem  of  §  42,  combined  with  Des- 
cartes' Rule  of  Signs  and  the  theorems  on  the  superior  and 
inferior  limits  of  the  roots  (§§  38-41),  Sturm's  Theorem  being 
used  only  when  the  other  theorems  fail  to  give  us  the  desired 
information. 

47.  Sturm's  Functions.     Let  f(x)  =  0  be  an  equation  which 
has  no  equal  roots.     Find  the  first  derived  function  of  f(x), 
namely  f'(x).     Then  proceed  with  the  process  of  finding  the 
highest  common  factor  of  f(x)  and  /'(#),  with  this  modification, 
that  the  sign  of  each  remainder  be  changed  beJforeJLiS-Msed  as  a 
tfivfsox.     Continue  the  process  until  a  remainder  is  reached 
which  does  not  contain  x,  and  change  the  sign  of  that  also.     We 
designate  the  several  remainders  with  their  signs  changed,  by 


LOCATION    OF    THE    ROOTS    OF   AN   EQUATION  51 

f2(x),  f3(x),  •'•,  fn(x),  and  call  them  auxiliary  functions.  The 
functions  /(ic),  f'(x),  f*(x},  fa(x)>  •",  /„(#)  are  called  Sturm's 
functions. 

&•  48.  Sturm's  Theorem.  If  f(x)  =  0  has  no  equal  roots,  let 
any  two  real  quantities  a  and  b  be  substituted  for  x  in  Sturm's 
functions,  then  the  difference  between  the  number  of  variations 
of  sign  in  the  series  when  a  is  substituted  for  x  and  the  number 
when  b  is  substituted  for  x  expresses  the  number  of  real  roots  of 
f(x)  =  0  between  a  and  b. 

When  f(x)  =  0  has  multiple  roots,  the  difference  between  the 
number  of  variations  of  sign  when  a  and  b  are  substituted  for  x 
in  the  series,  f(x)  ,  f  (x)  ,  f2(x)  ,  •  ••,  fr(x),  where  fr(x)  is  the  highest 
common  factor  of  f(x)  and  f'(x),  is  equal  to  the  number  of  real 
roots  between  a  and  b,  each  multiple  root  counting  only  once. 

First  Case.  No  Equal  Roots.  In  §  21  the  operation  of  find- 
ing the  highest  common  factor  between  /(#)  and/'(x)  was  used 
for  finding  multiple  roots  of  the  equation  f(x)  =  0.  If  there  is 
no  highest  common  factor  involving  x,  there  are  no  multiple 
roots,  and  we  are  able  to  find  all  of  the  n  +  1  Sturm's  functions. 
The  last  function,  fn(x),  is  numerical  and  not  zero. 

From  the  mode  of  formation  of  Sturm's  functions  we  obtain 
the  following  equations,  in  which  qi}  q2,  •••,  qn_i  are  the  succes- 
sive quotients  in  the  process  : 


/„-*<»  =  9»-i/»-i(*)  -/»(* 

(1)  Two  consecutive  auxiliary  functions  cannot  vanish  for 
the  same  value  of  x.  For,  if  f,(x}  and  fs(x~)  vanish  together 
when  x  =  c,  each  would  contain  the  factor  x  —  c.  From  the 
second  equation  it  would  follow  that  x  —  c  is  a  factor  of  /'  (x), 


52  THEOKY    OF    EQUATIONS 

and  from  the  first  equation  that  x  —  c  is  a  factor  of  f(x\ 
Hence  f(x)  and  f'(x)  would  have  a  common  factor  and  (§  21) 
f(x)  would  have  equal  roots,  which  is  contrary  to  hypothesis. 

(2)  When  any  auxiliary  function  vanishes  the  two  adjacent 
functions  have   opposite   signs.     Suppose,  for   example,  that 
f3(x)  is  zero  for  x  =  c.     By  (1),  f2(x)  and  /4(#)  cannot  be  zero 
when  fs(x)  is  zero.     The  third  equation,  above,  then  reduces 
to  f2(x)  =  —fi(x~),  showing  that  f2(x)  and  /4(#)  have  contrary 
signs. 

(3)  When  x,  in  passing  from  the  value  a  to  the  value  b> 
passes  through  a  value  which  makes  an  auxiliary  function 
vanish,  Sturm's  functions  neither  gain  nor  lose  variations  in 
sign.     For,  suppose  that,  for  x  =  c,  fr(x)  =  0,  then  /._i(c)  and 
fr+i(c)  have  opposite  signs.     As  fr(x)  passes  through  zero,  it 
changes  its  sign  from  +  to  —  ,  or  from  —  to  +.     Thus  the 
three  functions  /r_i(#),  fr(x),  fr+\(x)  will  have  one  variation  in 
sign  just  before  x  =  c  and  also   just   after   x  —  c.     In   other 
words,  no  matter  which  sign  is  placed  between  two  unlike 
signs,  we  have  only  one  variation.      Hence  no  variation  is 
either  gained  or  lost  among  Sturm's  functions. 

(4)  When  x,  in  passing  from  the  value  a  to  the  value  b, 
assumes  a  value  which  is  a  root  of  the  equation  f(x)  =  0,  then 
Sturm's  functions  lose   one  variation   in   sign.     By  Taylor's 
Theorem,  §  18, 

/(c  -  h}  -/(c)  =  -  hf'(c)  +    /"(c)  -  ..., 


• 

I* 

For  very  small  values  of  h  the  sign  of  the  right  member  of 
each  expansion  will  be  the  same  as  the  sign  of  its  first  term. 
If  /(re)  vanishes  for  x  =  c,  so  that/(c)  =  0,  and  if  f(c}  is  posi- 
tive, f(c  —  h)  is  negative  and  /(c  +  Ji)  is  positive.  That  is,  the 
signs  of  f(x)  and  f'(x)  will  be  —  h  just  before  x  =  c,  and  -f  + 


LOCATION   OF   THE   ROOTS   OF   AN   EQUATION  53 

just  after  x  =  c.  Thus  one  variation  in  sign  is  lost.  Iff'(c)  is 
negative,  then  /(c  —  Ji)  is  positive  and  /(c  +  h)  is  negative. 
That  is,  the  signs  of  f(x)  and/'(#)  will  be  -\  —  just  before 
x  =  c,  and  --  just  after  x  =  c.  Hence  a  variation  is  lost,  as 
x  passes  through  a  root  of  f(x)  =  0,  whether  /'(c)  is  positive 
or  negative. 

We  have  now  shown  that,  whenever  x,  in  passing  continu- 
ously from  a  to  6,  assumes  a  value  which  makes  one  or  more 
auxiliary  functions  vanish,  while  /(#)  does  not  vanish  for  that 
value,  no  variations  of  sign  are  gained  or  lost  among  Sturm's 
functions  ;  but  every  time  that  x  assumes  a  value  which  causes 
f(x)  to  vanish,  one  variation  is  lost.  Hence,  the  number  of 
variations  lost,  as  x  goes  from  the  real  value  a  to  the  real 
value  6,  is  equal  to  the  number  of  real  roots  of  f(x)  =  0  be- 
tween a  and  b. 

Second  Case.  Equal  Roots.  In  the  case  of  equal  roots  the 
f  unction  s/(x)  and  f'(x)  have  a  common  factor;  hence  the  last 
of  Sturm's  functions  is  not  a  numerical  constant,  as  before; 
this  last  function  is  now  the  highest  common  factor  of  f(x) 
and  /'(#).  Let  Sturm's  functions  be  f(x),f'(x),f2(x),  •••,fr(x'). 

If  x  passes  through  a  root  of  f(x)  =  0,  which  is  not  a  mul- 
tiple root,  then  the  reasoning  of  the  First  Case  still  holds. 

But  if  f(x)  =  0  has  the  multiple  root  r,  and  if  x  =  r,  we 
have  a  different  state  of  things;  consecutive  functions  will 
vanish  simultaneously.  Suppose  that  r  is  an  m-multiple  root, 
then  /(a!)  =  (x  _  r)»(a,  _  n)  (3.  _  r,)  ..  . 

and  f'(x)  =  m(x  —  r)m~l(x  —  r^  (x—r2)  ••• 

+  (a-  r)m(x-r2)(x-rs}  — 
+  (x  —  rf(x  —  rj(x  —  r8)  ... 


Divide  f(x)  and  /'(a)  by  their  H.  C.  F.  (a;  —  r)""1,  and  we  get 
two  functions  g(x)  and  gi(x).  We  notice  that  f(x)  and  g^x) 
have  no  common  factor  and  therefore  cannot  vanish  simul- 
taneously. Let  g'(x)  be  the  first  derived  function  of  g(x). 


54  THEORY    OF   EQUATIONS 

We  find  that  ^(a;)  differs  from  g'(x)  only  by  the  presence  in 
g\(x)  of  the  positive  coefficient  m.  If  x  =  r,  then  (ji(x)  and 
g'(x)  have  the  same  sign;  for,  g^r)  =  m(r  —  r^)(r  —  r2)  «••  and 
g'(r)  =  (r  —  ri)(r  —  r2)  •••.  They  have  like  signs  also  for 

05  =  ?-,,  or  r2,  •  •• 

We  may  therefore  find  the  situation  of  the  roots  of  g(x)  =  0 
by  taking  g(x)  and  g\(x)  as  the  first  two  of  Sturm's  functions 
and  forming  from  these  two  the  rest  of  them.  This  is  per- 
missible, since  by  applying  the  reasoning  of  the  First  Case  it 
may  be  shown  that  this  new  set  of  functions  possesses  the  two 
fundamental  properties  that  as  x  passes  from  a  to  6  no  varia- 
tions of  signs  are  gained  or  lost  when  an  auxiliary  function 
vanishes,  and  that  one  and  only  one  variation  is  lost  when 
g(x)  vanishes. 

The  number  of  variations  in  sign  will  always  be  the  same  for 

TJllG  SBT16S  ./» /"  \     _*M/\     _/» /   \  _*»/\ 

/(*),/'(*),  /2(»,   •••  /r(»), 

as  for  g(x),  g^x),  gs(x),  •••  gr(x).  • 

For,  corresponding  terms  of  the  two  series  of  functions  differ 
always  only  by  the  factor  (x  —  r)m~l,  so  that,  for  any  value  of 
x,  the  signs  of  the  terms  in  the  first  series  are  all  the  same  as 
those  of  the  second  series,  or  the  signs  are  all  unlike. 

Hence,  by  examining  the  variations  in  signs  of  the  first  series 
we  can  find  out  how  many  real  roots  of  the  equation  g(x)  =  0 
lie  between  a  and  b,  and  this  number  of  roots  is  the  same  as  the 
number  of  real  and  distinct  roots  of  the  equation  f(x)  =  0 
between  those  same  limits.  This  proves  the  second  case  when 
r  is  a  multiple  root.  If  f(x)  =  0  has,  besides  r,  the  multiple 
root  rm,  then  a  slight  and  obvious  modification  of  our  proof  is 
necessary. 

49.  In  the  application  of  Sturm's  theorem,  the  following 
point  must  be  borne  in  mind.  In  finding  the  functions  /2(#), 
f3(x),  "•  it  is  allowable  to  introduce  or  suppress  any  monomial 


55 

or  numerical  factor,  as  is  done  in  the  process  of  finding  the 
H.  C.  F.,  provided  that  the  factor  is  positive.  Particular  care 
must  be  taken  not  to  change  any  of  the  signs,  except  of  course 
the  sign  of  a  remainder,  just  before  it  is  used  as  a  divisor  in  the 
next  operation. 

If  we  wish  to  ascertain  simply  the  total  number  of  real  roots, 
without  fixing  their  location,  we  need  only  substitute  in  the 
Sturmian  functions  the  values  x  =  —  oo  and  x  =  -f  oo  and 
observe  the  difference  in  the  number  of  variations  of  sign. 

Ex.  1.   Apply  Sturm's  Theorem  to  x3  -  x2  -  10  x  +  1  =  0. 
Here  /'(x)  =  3  x2  -  2  x  -  10, 

/2(x)  =  62  x  +  1, 

/3(x)  =  38313. 
We  give  the  signs  of  the  Sturm's  functions  for  the  indicated  values  of  x  : 

oo  +  +  +  + 

4  +  +  +  + 

31  I  I 

•*—  -p  -f-  -f-  ,     £^^  ^      J  L|        / 

1  +  + 

0  +  -*•  +  +          I    Vvl 

-2  + "          +  + 

-3  +  + 

-co  +  + 

Since  x  =co  gives  no  variations  and  x  =  —  co  gives  three  variations,  all 
three  roots  are  real.  The  roots  lie  between  3  and  4,  0  and  1,  —  2  and  —  3. 

Ex.  2.  Apply  Sturm's  Theorem  tox5-5x4  +  9x8-9x2  +  6x-l  =  0, 
the  equation  given  in  Ex.  2,  §  42. 

Here  /'(x)  =  5x*  -  20  x3  +  27  x2  -  18  x+  5, 

/3(x)  =  -  32  x2  +  38  x  -  6,         X.    /y\   -    -L— 
f>     .  /      \     vA  1        j 

f  ( rr\    9ft  T.     i     1Q  ^» 

J±\*>)    —    —  6V  *  T   *W,  ^ 

/5(x)  =  -  192. 

When  x  =00  ,  Sturm's  functions  give  one  variation  ;  when  x  =  — oo,  they 
give  four.  Hence  there  are  three  real  and  two  imaginary  roots. 


56  THEORY    OF    EQUATIONS 

Ex.  3.   Apply  Sturm's  Theorem  to  2  x5  +  1  x*  +  8  x3  +  2  x2  -  2  x  -  1  =  0. 
We  find  /'(x)  =  10z4  +  28  x3  +  24  x2  +  4  x  -  2, 

/2(x)  =  x*  +  3  x2  +  3  *  +  1. 

Here  /2(x)  is  found  to  be  the  H.  C.  F.  of  /(x)  and  /'(a;);  hence  -  lisa 
quadruple  root.  For  x  =  +00  ,  the  functions  /(«),  /'(#),  /a(x)  yield  the 
signs  +  +  +  ;  for  x=  -co  they  yield  —  |  --  .  Hence  there  are  two 
distinct  real  roots,  and  all  the  roots  are  real. 

Ex.  4.  Show  that  all  the  roots  of  x4  +  Xs  —  x2  —  2  x  +  4  =  0  are 
imaginary. 

Ex.  5.  Required  the  number  and  situation  of  the  real  roots  of 

2z*-  Ilx2  +  8x-16=0, 
X3+  Iix2-102x+  181  =  0, 
x5  _  36  X3  +  72  X'2  -  37  x  +  72  =  0. 

50.  Nature  of  the  Roots  of  the  Quartic.  In  the  study  of  the 
nature  of  the  roots  of  the  cubic  equation  we  began  in  §  35  by 
deducing  the  "  equation  of  squared  differences  of  the  roots  of 
the  cubic."  Then,  in  §  36,  we  used  this  transformed  equation 
in  the  discussion  of  the  roots  of  the  given  cubic.  The  same 
mode  of  procedure  might  be  adopted  in  the  study  of  the  roots 
of  the  quartic  equation.  But  the  formation  of  the  "  equation 
of  squared  differences  of  the  roots  "  is  laborious,  and  we  prefer 
to  begin  the  discussion  by  applying  Sturm's  Theorem  to  the 
quartic  with  its  second  term  removed. 

If  we  transform  the  general  quartic 


into  a  new  equation,  deprived  of  its  second  term  and  with  coef- 
ficients integral  in  form,  we  obtain,  as  in  §  34, 


where 


=  6064  _  4  ifa  _|_  3  b/. 


LOCATION   OF   THE   ROOTS   OF   AN    EQUATION  57 

Representing  the  left  member  of  equation  II  by  f(y),  we  get 


and,  by  division,    <=•      •         A^*v""^ 

-3Hy*-3Gy  -bjl+  3  H*. 


Before  dividing  \f'(y)  by  f2(y),  multiply  \f'(y)  by  the  posi- 
tive factor  3  H2.     We  obtain,  after  dividing  •  the  remainder 

by  V'         My)  =  (Vtf/-  3  G*  -  12  £T3)  £  -  <?/. 

00 

We  find  it  convenient  to  let  b02HI-  G2-±HS  =  b<?J. 


Then         /a(y)  =  (3  &„/-  2  JZ7>  -  GL 

Now  multiply  /2(y)  by  the  positive  factor  (3  b0J  —  2  HI)2, 
and  we  obtain,  after  division,  a  remainder  which,  with  its  sign 
changed,  is  equal  to 

(&02/-  3  #2)(3  b<>J-  2  HI)2  +  3  G2I(3  bvl-  HI) 

=  bJH*!*  -  27  602^"2./2  +  T, 
where         T=  (9  b<?IJ2-I2  b0sHI2J+36  b0HsIJ+3b0G2IJ) 

+  (3  fcolEPr5  -  3  G2I2H-  12  IT4/2) 
=  3  60/«/(3  6bV-4  b02HI+12  H3+3  G2)+3  b03I2HJ 
=  3  b0IJ(3  b03J-  3  b02HI+  12H3  +  3  G2) 
=  3  boIJ(3  b03J-  3  b<?J)  =  0. 

If  the  remainder  is  divided  by  the  positive  factor  b^H2,  we 
obtain  f4(y-)  =  I*-27J2. 

We  have  now  all  of  Sturm's  functions  of  equation  II. 

(1)  All  roots  real.  If  (73  -  27  J2)  >  0,  (3  bvT  -2HI)>  0, 
and  H<  0  ;  then,  for  y  =  &>,  the  signs  of  Sturm's  functions  are 
+  -f  +  +  -f  ;  for  y  =  -  oo  the  signs  are  -\  ---  1  ---  h-  The 
excess  of  variations  in  the  latter  case  is  four  ;  hence  all  the 
roots  are  real. 


58  THEORY   OF    EQUATIONS 

(2)  All  roots  imaginary.     If  /3-27.72>0,  and  if  H  >0  or 
else  (3  bqJ—  2  HI)  <  0,  then  the  number  of  variations  in  signs 
for  y  =  —  oo  is  the  same  as  for  y  =  oo ;  hence  there  are  no  real 
roots. 

(3)  Two  real  roots.     If  Is  —  27  ,72  <  0,  then,  no  matter  what 
signs  H  and  (3  b^J—  2  HI)  may  have,  we  get  always  a  dif- 
ference of  two  variations  for  y  =  cc  and  y  =  —  oo ;  hence  there 
are  two  real  roots  and  two  imaginary  roots. 

(4)  Equal  roots.    When  Is  —  27  J2  =  0,  it  is  evident  from  the 
theory  of  the  H.  C.  F.  that  there  are  equal  roots.    If  f+(y)  is  the 
only  one  of  Sturm's  functions  which  vanishes  identically,  then 
/3(y)  is  the  H.  C.  F.  in  y  and  there  are  two  roots  equal  to  each 
other.     If  fs(y)  is  identically  zero,  which  happens  when  7=  0 
and  J=  0,  or  when  G  =  0  and  3  b^T—  2  HI,  then  three  roots  are 
equal  to  each  other  or  there  are  two  distinct  pairs  of  double 
roots.     That  is,  if  7=  0  and  J=  0,  we  get  from  the  equation 
defining  J  the  relation  G"  +  4  Hs  =  0,  which  makes  f2(y)  a  per- 
fect square.     Hence  three  roots  are  equal.     When  G  =  0  and 
3  b0J=  2  HI,  it  follows  that  602/=  1.2  H2  and/2(?/)  is  readily  seen 
to  be  composed  of  two  unequal  factors  in  y,  indicating  the  ex- 
istence of  two  distinct  pairs  of  equal  roots.     If  we  have  7=  0, 
,7=  0,  and  H=  0,  then  it  follows  that  G  =  0  and/2(y)  =  0;  hence 
f(x)=y3  and  all  the  roots  are  equal. 

This  discussion  of  equation  II  applies  also  to  equation  I, 
representing  the  general  quartic  ;  for,  since  y  —  b<&  -f  blt  the 
values  of  x  are  real,  imaginary,  or  multiple  values,  according 
as  the  values  of  y  are  real,  imaginary,  or  multiple  values. 

Ex.  \1.   Compute  the  values  of  H,  G,  /,  J  for  the  equation 

re*  -  4z»  +  60  z2  -  8  a;  +  1  =  0. 
Then  discuss  the  nature  of  the  roots. 

Ex.  2.  Show  that  in  equation  II  a  double  root  is  equal  to  GI  -*- 
(3  b0J—2  HI},  a  triple  root  is  equal  to  —  iff2,  a  quadruple  root  is  equal  to  0. 

Ex.  3.  Apply  Sturm's  Theorem  to  the  cubic  yz  +  3  Hy  +  G  =  0,  and 
verify  the  results  of  §  36. 


LOCATION   OF   THE   BOOTS   OF   AN   EQUATION          59 

51.  Discriminant  of  the  Quartic.  The  expression  Is—  27  J2 
played  an  important  role  in  the  discussion  of  the  nature  of  the 
roots  of  the  quartic.  We  shall  prove  that,  when  multiplied  by 
the  constant  256  60~6,  it  is  equal  to  the  product  of  the  squares 
of  the  differences  of  the  roots.  This  product  is  called  the 
discriminant  of  the  quartic. 

Let  I3  —  27  J2  =  R.    When  R  vanishes,  the  quartic  was  seen 

r    ft"* 

to  haVe  equal  roots.  Hence  («  —  «i)  must  be  a  factor  of  R. 
Since  J?  is  a  constant  for  an  equation  with  constant  coefficients, 
it  is  unaltered  when  («  —  «i)  is  changed  to  («!  —  «).  Hence 
(a  —  ttj)2  must  be  a  factor  of  R.  This  reasoning  holds  for  the 
difference  of  every  two  roots.  Hence 


••(«2-a3)2,  I 

is  a  factor  of  7?.  Remembering  that  b1}  b2,  63,  64  are  symmetric 
functions  of  the  roots,  involving  the  roots  to  the  degrees  one, 
two,  three,  four,  respectively,  we  see  on  examining  the  expres- 
sion for  R,  that  it  cannot  involve  products  of  roots  of  higher 
degree  than  12.  But  12  is  also  the  degree  of  the  terms  in 
the  product  I.  Hence  there  are  no  other  factors  in  R  which 
involve  the  roots.  Therefore,  R  differs  from  the  product  I  by 
some  numerical  factor  only.  This  factor  can  be  easily  found 
by  using  any  simple  quartic  which  has  distinct  roots,  say 
b^  —  1  =  0.  Here  R  =  —  603,  the  product  I  is  —  256  60~3-  Hence 

(a  —  <*i)2(«  —  «2)2(a  ~~  as)2(ai  —  a2)2(ai  —  °£3)2(()C2  ~ 

\    : 
where  D  is  the  discriminant. 


CHAPTER  IV 

APPROXIMATION   TO  THE   ROOTS   OF  NUMERICAL 
EQUATIONS 

52.  Solution  by  Radicals  and  by  Approximation.  The  modern 
theory  of  equations  is  the  outgrowth  of  attempts  made  during 
past  centuries  to  solve  equations  arising  in  the  consideration 
of  problems  in  pure  and  applied  mathematics.  The  subject  of 
the  solution  of  equations  resolves  itself  into  two  quite  distinct 
parts :  Firstly,  the  solution  of  numerical  equations  whose 
coefficients  are  given  numbers,  by  some  method  of  approxima- 
tion to  the  true  value  of  the  roots  ;  secondly,  the  solution  of 
equations  whose  coefficients  are  either  particular  numbers  or 
independent  variables,  in  such  a  way  as  to  yield  accurate  expres- 
sions for  the  values  of  the  roots  in  terms  of  the  coefficients  — 
such  expressions  to  involve  no  other  processes  than  addition, 
subtraction,  multiplication,  division,  and  the  extraction  of  roots 
of  any  orders.  The  latter  process  is  called  the  algebraic  solution 
of  equations.  The  former  is  of  importance  to  the  practical 
computer,  the  latter  is  of  special  interest  to  the  pure  mathema- 
tician. In  the  former  each  root  may  be  determined  separately ; 
in  the  latter  a  general  expression  must  be  found  which  repre- 
sents all  the  roots  indifferently. 

In  the  algebraic  solution  of  equations  no  great  difficulty 
presents  itself  as  long  as  the  degree  of  the  equation  does  not 
exceed  four.  But  in  spite  of  persistent  attempts  by  many  of 
the  ablest  mathematicians,  no  algebraic  solution  of  the  general 
equation  of  the  fifth  or  a  higher  degree  has  ever  been  given. 
In  fact,  we  shall  be  able  to  show  conclusively  that  no  such 
solution  is  possible ;  that  is,  no  solution  can  be  given  in  which 

60 


BOOTS   OF   NUMERICAL   EQUATIONS  61 

the  roots  are  expressed  in  terms  of  the  coefficients  by  means 
of  radical  signs  or  fractional  exponents.  In  the  quadratic 
a?  +ax  -+-  b  =0  we  know  that  x  =  ^  ( —  a  ±  V«2  —  46).  In  the 
cubic  we  shall  see  that  x  can  be  similarly  expressed  in  terms 
of  its  coefficients  by  indicating  the  extraction  of  certain  square 
roots  and  cube  roots.  The  same  remark  applies  to  the 
quartic.  But  in  the  general  quintic  x  refuses  to  submit  itself 
to  this  mode  of  treatment.  A  general  solution  of  the  quintic 
has  been  given,  but  the  solution  involves  elliptic  integrals 
and  is,  therefore,  not  algebraic,  but  transcendental. 

The  problem,  of  the  solution  of  numerical  equations  by 
approximation  to  a  certain  number  of  decimal  ^Zaces  is  much 
easier.  Not  only  are  we  able  to  determine,  with  comparative 
ease,  the  real  roots  of  equations  of  lower  degrees,  but  also  of 
the  quintic  and  of  higher  equations. 

Methods  of  approximation  to  the  roots  of  numerical  equa- 
tions have  been  devised  by  several  mathematicians  —  Newton, 
Lagrange,  Biidan,  Fourier,  and  others.  But  the  best  practical 
method  is  that  given  in  1819  by  William  George  Homer. 
We  shall  confine  ourselves  to  the  exposition  of  his  method 
and  that  of  Newton. 

/  ~  53.  Commensurable  and  Incommensurable  Roots.  A  real  root 
of  a  numerical  equation  is  said  to  be  commensurable  when  it  is 
an  integer  or  a  rational  fraction ;  it  is  said  to  be  incommensur- 
able when  it  involves  an  interminable  decimal  which  is  not  a  re- 
peating decimal..  Since  a  repeating  decimal  can  be  expressed 
as  a  rational  fraction,  a  root  in  that  form  is  commensurable. 

54.  Fractional  Roots.  A  rational  fraction  cannot  be  a  root  of 
an  equation  with  integral  coefficients,  the  coefficient  of  xn  being 
unity. 

If  possible,  let  -,  h  and  Jc  being  integers  and  -  a  fraction 
k  7c 

reduced  to  its  lowest  terms,  be  a  root  of  the  equation 
xn  +  a^-1  -f  a^-2  -\ H  an  =  0. 


62  THEORY   OF   EQUATIONS 

Writing  -  for  x,  we  get 

K 


Multiplying  by  A;""1  and  transposing  all  integral  terms, 

—  —  _  aJin-1-  aJin~2k  -----  aJfc—1. 
k 

hn 
This  equation  is  impossible,  since  the  fraction  —  ,  which  is  in 

K 

its  lowest  terms,  cannot  be  equal  to  an  integral  number. 
Hence,  -  cannot  be  a  root  of  the  given  equation. 

K 

~o    55.   Integral    Roots.      Since    the    equation    with    integral 
coefficients,  xn  +  aiJC«-i  +  .  .  .  +  ^  =  0, 

cannot  have  rational  fractional  roots,  and  since  an  is  numerically 
equal  to  the  product  of  all  the  roots  (§  13),  it  is~evident  that  all 
commensurable  roots  are  exact  divisors  of  an  and  may  be  found 
by  testing  the  factors  of  an.  By  §  4  a  factor  c  is  a  root,  if 
f(x)  is  divisible  by  x  —  c  without  a  remainder. 

If  the  coefficient  of  xn  is  not  unity,  but  a0,  then  we  may 
divide  through  by  OQ  and  transform  the  equation  into  another 
whose  roots  are  those  of  the  given  equation  multiplied  by  «0 
(§  29).  In  the  new  equation  the  coefficient  of  xn  is  unity  and 
all  the  other  coefficients  are  integral.  Hence,  all  its  commen- 
surable roots  are  integral. 

Ex.  1.     Find  the  commensurable  roots  of  x3  —  7  x  —  6  =  0. 

The  commensurable  roots  must  be  found  among  the  values  ±1,  ±2, 
±3,  ±6,  which  are  all  factors  of  —  6.  By  Descartes'  Rule  of  Signs  we 
see  that  there  is  only  one  positive  root.  By  substitution  or  by  synthetic 
division  we  find  that  +  1  is  not  a  root,  that  —  1  is  a  root.  We  may  now 
either  depress  the  degree  of  the  equation  by  dividing  by  x  +  1  and  then 
solve  the  resulting  quadratic,  or  we  may  try  the  other  factors.  We  obtain 
—  2  and  +  3  as  the  values  of  the  other  roots. 

Ex.  2.     Find  the  commensurable  roots  of 

2  x8  -  x2  -  x  -  3  =  0. 


HOOTS   OF   NUMERICAL   EQUATIONS  63 

Dividing  the  left  member  by  2  and  multiplying  the  roots  by  2,  we 
obtain  x3  -  x2  -  2  x  -  12  =  0. 

It  is  found  that  +  3  is  the  only  commensurable  root  of  this  equation. 
Hence,  +  f  is  the  only  commensurable  root  of  the  given  equation. 

Ex.  3.     Find  all  the  commensurable  roots  of 
x3  +  4x2  +  6x  +  3  =  0. 
x4  -  3  x8  -  22  x2  -  39  x  -  21  =  0. 
&  _  10  tf  +  17  x2  -  x  -  7  =  0. 
x5  -  13x*  +  34 x3  -  26 x2  -  18x  +  22  =  0. 
6  x3  -  25  x2  +  3  x  +  4  =  0. 
4  x3  +  20  x2  -  23  x  +  6  =  0. 

"O  56.  Hornet's  Method.  This  method  may  -be  used  advanta- 
geously for  finding  not  only  incommensurable  roots,  but  also 
commensurable  roots  when  the  process  of  §  55  is  inconvenient. 

In  the  application  of  Homer's  method  we  must  know  the 
first  significant  figure  of  the  root,  to  start  with.  The  first  digit 
may  be  found  by  the  process  indicated  in  §  42  or  by  Sturm's 
Theorem. 

Homer's  method  consists  of  successive  transformations  of  an 
equation.  Each  transformation  diminishes  the  root  by  a  certain 
amount.  If  the  required  root  is  2.24004,  then  the  root  is 
diminished  successively  by  2,  .2,  .04,  .00004.  The  mode  of 
effecting  these  transformations,  by  synthetic  division,  was 
explained  in  §  32.  The  method  will  be  readily  understood  by 
the  study  of  the  following  example : 

Ex.  1.    The  equation      x3  -  x  -  9  =  0,  I 

has  a  root  between  2  and  3,  for  /(2)  =  -  3  and  /(3)  =  +  15.  The  first 
figure  of  the  root  is  therefore  2.  Transforming  the  equation  so  that  the 
roots  of  the  new  equation  will  be  smaller  by  2,  we  obtain 


+  0 

-  1 

-9  [2 

+  2 

+  4 

+  6 

+  2 

+  3 

-3 

+  2 

+  8 

+  4 

+  11 

+  2 

+  6 

64 


THEORY   OF   EQUATIONS 


Since  the  roots  of  the  transformed  equation 

x3  + 6x2  + llx-3  =  0  II 

are  equal  to  the  roots  of  equation  I  less  2,  equation  II  has  a  root  between 
0  and  1.  This  root  being  less  than  unity,  x2  and  z3  are  each  less  than  x. 
Neglecting  x3  and  6  x2,  we  obtain  an  approximate  value  for  x  from 

11  x  -  3  =  0,  or  x  =  .2. 
Transforming  II  so  as  to  diminish  the  roots  by  .2,  we  get 

x3  +  6.6  x2  +  13.52  x  -  .552  =  0.  Ill 

Neglecting  x3  +  6.6  x2,  we  find  an  approximate  value  for  x  in  equation 
-II  from  13.52  x-. 552  =  0,  or  x  =  .04. 


Diminishing  the  roots  of  III  by  the  value  .04,  we  have 
x3  +  6.72  x2  +  14.0528  x  -  .000576  =  0. 


IV 


From  14.0528  x  -  .000576  =  0,  we  get  x  =  .00004. 

The  root  of  equation  I  whose  first  figure  is  2  has  now  been  diminished 
by  2,  .2,  .04,  .00004.  Hence  the  root  is  approximately  2.24004.  The  suc- 
cessive transformations  may  be  conveniently  and  compactly  represented 
as  follows : 


+  0 
2 


2.448 


6.6 

.04 
6.64 

.04 
6.68 

.04 
6.72 


14.0528 


4 
2 

11 
1.24 

-.552 
.551424 

6 

.2 

12.24 
1.28 

-  .000576 

6.2 

.2 

13.52 
.2656 

6.4 
.2 

13.7856 
.2672 

ROOTS   OF   NUMERICAL   EQUATIONS 


65 


The  broken  lines  indicate  the  conclusion  of  the  successive  transforma- 
tions. The  numbers  immediately  below  a  broken  line  are  the  coefficients 
of  the  transformed  equation.  Thus,  the  second  transformed  equation  is 
seen  at  once  to  be  x3  +  6.6  x2  +  13.52  x  —  .552  =  0. 

Ex.  2.  In  the  equation  x3  —  46.6  x2  —  44.6  x  —  142.8  =  0  we  find  that 
/(40)  =  — ,  /(50)  =  + .  Hence  there  is  a  root  between  40  and  50.  To  find 
this  root,  diminish  the  roots  by  40,  then  find  the  first  figure  of  the  root  hi 
the  transformed  equation  and  proceed  by  Homer's  method  as  already 
explained.  The  work  is  as  follows : 


-46.6 
40 

-44.6 
-264 

-  142.8 
-  12344 

-6.6 
40 

-  308.6 
1336 

-  12486.8 
11131.4 

33.4 
40 

1027.4 

562.8 

-  1355.4 
1355.4 

73.4 

7 

1590.2 
611.8 

80.4 

7 

2202.0 
57 

87.4 

7 

2259 

94.4 
.6 

95 

In  the  first  transformed  equation  x3  +  73.4  x2  +  1027.4  x  -  12486.8  =  0 
we  only  know  that  the  value  of  x  is  less  than  10 ;  hence  the  method  of 
Ex.  1,  where  we  ignored  the  terms  containing  x3  and  x2,  is  not  applicable. 
Since  in  this  transformed  equation  /(7)  =  —  and  /(8)  =  + ,  we  know  that 
7  is  the  desired  digit. 

In  the  second  transformed  equation  we  know  that  x  lies  between  0  and 
1.-  Hence  we  find  the  first  digit  of  x  from  the  equation  2202  x  — 1355.4=0. 

Since  in  the  third  transformation  there  is  no  remainder,  we  know  by 
§  3  that  .6  is  a  root  of  x3  4-  94.4  x2  +  2202  x  -  1355.4  =•  0  and  that  47.6  is  a 
commensurable  root  of  the  given  equation. 

When  the  fractional  part  of  the  root  is  being  found  and  the 
values  of  the  coefficients  of,  y?,  etc.,  are  sufficiently  small,  it  will 
be  noticed  that  the  last  two  terms  of  each  transformed  equation 
occurring  in  Horuer's  process  have  opposite  signs.  This  is  as  it 


66  THEORY   OF   EQUATIONS 

should  be ;  for  if  the  two  terms  had  like  signs,  the  value  of  x  in 
the  transformed  equation  would  be  negative,  showing  that  the 
last  digit  in  the  root  of  the  original  equation  had  been  taken  too 
large.  For  instance,  if  in  Ex.  1  the  first  decimal  had,  by  mis- 
take, been  taken  as  3,  instead  of  2,  then  the  second  trans- 
formed equation  would  have  been  or3  -f  6.9  x2  +  14.87  x  +  .867  =  0. 
The  approximate  value  of  x  in  this  equation  is  —  .05,  show- 
ing that  in  diminishing  the  roots  by  .3  we  took  away  too 
much. 

If,  by  mistake,  a  digit  is  taken  too  small,  the  error  will  show 
itself  in  the  next  step.  Suppose  that  in  Ex.  1  the  first  deci- 
mal had  been  taken  to  be  .1,  then  the  second  transformed 
equation  would  have  been  ar5  +  6.3  3*  -f  12.23  x  —  1.839  =  0. 
From  12.23  x  — 1.839  =  0  we  get  approximately  x  —  .15.  This 
changes  .1  into  .25,  and  thus  discloses  an  error  in  the  estimate 
of  the  first  decimal. 

To  find  the  value  of  a  negative  root  by  Horner's  method,  we 
need  only  transform  the  given  equation  by  writing  —  x  for  x 
and  then  proceed  as  before. 

Ex.  1.   Find  the  real  roots  of : 

(1)  4  x5  -  3  x1  -  2  x2  +  4  x  -  10  =  0. 

(2)  3  x5  +  3  x*  -  x2  -  4  x  +  5  =  0. 

(3)  7  x*  +  3  x3  -  5  x2  +  4  x  -  6  =  0. 

(4)  x"  -  x6  +  x5  +  x*  -  10  =  0. 

(5)  z5  -  4  x  -  2  =  0. 

^  57.  Newton's  Method  of  Approximation.  This  method  is  not 
as  convenient  in  the  solution  of  numerical  equations  involving 
algebraic  functions  as  is  the  method  of  Horner,  but  it  has  the 
advantage  of  being  applicable  to  numerical  equations  involv- 
ing transcendental  functions.  For  instance,  Newton's  method 
can  be  used  in  finding  x  in  x  —  sin  x  =  2. 

Let  f(x)  =  0  be  the  given  equation.  Suppose  that  we  know  a 
quantity  a  which  differs  from  one  of  the  values  of  x  by  the  small 


ROOTS   OF   NUMERICAL   EQUATIONS  67 

quantity  Ti.     Then  we  have  x  =  a  +  h.     By  Taylor's  Theorem 
f(x)  =f(a  +  Ji)  =f(a)  +  hf(a)  +| /"(a)  -f  .... 

Since  h  is  small,  we  get,  by  neglecting  higher  powers  of  h,  an 
approximate  value  of  h  from  the  equation  /(a)  +  hf'(a)  =  0, 

namely,  h  =  —  •{,;  (  *    We  have  approximately  »  =  a  —  J(a\ 
/  («)  /'(a)  • 

Letting  this  new  approximation  to  the  value  of  x  be  repre- 
sented by  6,  we  may  repeat  the  above  process  and  secure  a 
still  closer  approximation,  and  so  on. 

Ex.  1.   Solve  x  —  sin  x  =  2. 

The  angle  x,  measured  in  radians,  must  lie  between  2  and  3.  Take 
a  =  2.5,  .j.^  _  5  _  sin  2.5  -  .5  _  sin  1430  14/  _  _  -097. 

/'(a)  =  1  -  cos  2.5  =  1.801. 
Hence  h  =  .0539,  b  =  a  +  h  =  2.5539. 

A  second  approximation  gives  us 

/(&)  =  -.00054,  /'(&)  =>  1.8322,  A=  .0002947. 
Hence  x  =  b  +  h  =  2.554195. 

58.  Complex  Roots  of  Numerical  Equations.  Recently  methods 
for  approximating  to  the  complex  as  well  as  the  real  roots  of 
numerical  equations  have  been  perfected.*  The  exposition  of 
these  methods  is  too  long  for  a  work  like  this. 

*  See  Emory  McClintock,  "  A  Method  for  Calculating  Simultaneously  All  the 
Roots  of  an  Equation,"  in  the  American  Journal  of  Mathematics,  Vol.  XVII., 
pp.  89-110  ;  M.  E.  Carvallo,  Mtthode  pratique  pour  la  Resolution  numtrique 
complete  des  Equations  algebriques  ou  transcendantes,  Paris,  1896. 


CHAPTER  V 

THE  ALGEBRAIC   SOLUTION   OF   THE   CUBIC  AND   QUARTIC 

59.    Solution  of  the  Cubic.    There  are  many  different  solutions 
of  the  general  cubic  equation, 


The  one  which  we  shall  give  is  due  to  the  Italian  mathematician 
Tartaglia  and  was  first  published  in  1545  by  Cardan.  Equa- 
tion I  is  first  transformed  into  another  whose  second  term  is 

wanting.     Putting,  as  in  §  33,  x  =  z~   1,  we  get 

' 


where  H  =  b0b2  -  bf  and  G  =  b02b3  -  3  60&i&2  +  2  bf.     To  solve 
equation  II,  let  z  =  u  +  v.     Substituting  in  II,  we  get 

u3  +  v3  +  3(uv  +'H)(u  +  v)  +  G  =  0. 

We  are  permitted  to  subject  the  quantities  u  and  v  to  a  second 
condition.     The  most  convenient  assumption  will  be 

uv  +  H=  0.  Ill 

This  yields  us  +  v3  =  -  G.  IV 

Eliminating  v  between  III  and  IV,  we  get 

TT3 

us-  —  =  -G,  or  u«+Gus  =  H*. 
u3 

The  last  equation  is  a  quadratic  in  form.     Solving  it,  we  have 


68 


SOLUTION   OF   THE   CUBIC    AND    QUARTIC  69 


Then  by  IV,         vs  =  -  G  -  us=  -®-^^+  H3. 


ri 

Since    U  =      -     + 


and  z  =  u  +  v,  we  have 


The  expression  for  the  root  of  the  cubic,  given  in  formula  VI 
is  known  as  Cardan's  formula. 

Since  a  number  has  three  cube  roots,  it  is  evident  from  V 
that  u  and  v  have  each  three  values.  It  may  seem  as  if  with 
-  each  value  of  u  we  might  be  able  to  associate  any  one  of  the 
three  values  of  v,  thus  obtaining  all  together  nine  values  for 
u  +  v,  or  z.  As  the  cubic  has  only  three  roots,  this  cannot  be. 
Of  the  nine  values,  six  are  excluded  by  equation  III,  which  u 
and  v  must  satisfy.  Eliminating  v  between  z  =  u  +  v  and  equa- 
tion III,  we  get  TT 

z  =  u-S-,  VII 

u 

where  u  has  the  form  given  in  V.  Since  in  expression  VII  there 
is  only  one  number,  u,  which  has  triple  values,  this  expression 
does  not  involve  the  difficulties  of  Cardan's  formula.  Let  the 
three  values  of  u  be  u,  u<a,  uu>2,  where  o>  stands  for  one  of  the  two 
complex  cube  roots  of  unity,  —  £±|-V—  3.  Then  the  three 
roots  of  the  cubic  II  are 

H  Hu?  2         Hut  -I7TTT 

u ,  wo) ,  uur —  • •  Vlli 

u  u  u> 

Since  z  =  b^c  +  bl}  we  obtain  the  roots  of  the  general  cubic  I 
by  subtracting  ^  from  each  of  the  three  expressions  in  VIII, 
and  then  dividing  the  three  results  by  b0. 

60.  Irreducible  Case.  —  The  general  expression  for  the  roots 
of  a  quadratic  equation  with  literal  coefficients  may  be  used 


70  THEORY   OF    EQUATIONS 

conveniently  in  solving  numerical  quadratic  equations.  For 
each  letter  we  substitute  its  numerical  value,  then  carry  out 
the  indicated  operations.  It  is  an  interesting  fact  that,  in  case 
of  the  cubic,  this  mode  of  procedure  is  not  always  possible  and 
that  the  algebraic  solution  of  the  cubic  is  of  little  practical  use 
in  finding  the  numerical  values  of  the  roots. 

In  §  36  we  found  that  the  roots  of  the  cubic  are  all  real 
when  G2  -\-  4  H3  is  negative.  In  the  attempt  to  compute  these 
real  roots  of  the  cubic  by  substituting  the  values  of  H  and  G 
in  the  general  formula,  we  encounter  the  problem,  to  extract 
the  cube  root  of  a  complex  number.  But  there  exists  no  con- 
venient arithmetical  process  of  doing  this.  Nor  is  there  any 
way  of  avoiding  the  complex  radicals  and  of  expressing  the 
values  of  the  real  roots  by  real  radicals.  This  fact  will  be 
proved  in  Ex.  8,  §  183.  By  the  older  mathematicians  this 
case,  when  G2  +  4  H3  is  negative,  was  called  the  "  irreducible- 
case  "  in  the  solution  of  the  cubic,  the  word  "  irreducible  "  hav- 
ing here  a  meaning  different  from  that  now  assigned  to  it  in 
algebra.  See  §  123. 

61.  Solution  by  Trigonometry.  The  "irreducible  case"  may 
be  disposed  of  by  expanding  the  two  terms  in  Cardan's  formula 
into  two  converging  series  with  the  aid  of  the  binomial  theorem. 
The  imaginary  terms  will  disappear  in  the  addition  of  the  two 
series.  But  it  is  better  to  use  the  following  trigonometric 
method  (which  is  itself  inferior,  for  the  purpose  of  arithmetical 
computation,  to  Homer's  method,  §  56) : 


Let 

We  get  us  =  r  (cos  6  +  i  sin  0), 

v*  =  r  (cos  0  —  i  sin  6), 

where 


SOLUTION    OF   THE   CUBIC    AND   QUARTIC 


71 


=  V  —  ±±     COS 


Hence, 


and 


where  w  takes  the  values  0,1,2. 


+  ?'sin- 


3      / 


62.   Euler's  Solution  of  Quartic.     Eemoving  the  second  term 
of  the  quartic 


we  get  as  in  §  34, 


where    z=b<ps  +  61}  £f=  60&2  —  V,  -*"=  &o&4  —  4  b:ba  +  3  622, 
(?  _=  60263  _  3  &0&A  +  2  &!». 

Euler  assumes  the  general  expression  for  a  root  of  equation  II 


z  = 

Squaring,  z2  —  it  —  -y  —  1«  =        w     v 
Squaring  again  and  simplifying, 

z*  —  2zz(u  +  v  +  w)—Sz  ~Vu  ~Vv  -\fw  +  (u  +  v  +  w) 
—  4  (wu  +  uw  -\-  vw)  =  0. 

Equating  coefficients  of  this  and  equation  II,  we  have 


(u  +  v  -\-  wY  —  4  (uv  +  uw 
or  uv  +  uw  +  vw  = 


But  —  (w  +  v  +  w),  (wv  +  uw  +  vw),  —  ?<uw  are  the  coefficients 
of  a  cubic  whose  roots  are  u,  v,  w.     This  cubic,  called  "  Euler's 

Cubic,"  is 

Ill 


=  602-^  —  3  H2, 


72  THEOKY   OF   EQUATIONS 

Let  y  =  b02x  —  H,  and  we  obtain 


0,  IV 

where  b0sJ=  b02IH-  4  S3  -  G\ 

Equation   IV  is  called  the  reducing  cubic  of  the   quartic. 
Since  u,  v,  w  are  the  three  values  of  y  in  III,  we  have 

u  =  bi2  —  b0b2  +  fyfco    v  =  b?  —  b0b2  +  b02x2)  w  =  bt2  —  b0b2  +  b02xs. 
Hence, 


Or,  since  G  =  —  2     w     v     w,  we  may  write 


In  the  expression  for  z  in  VI  each  of  the  radicals  may  be 
either  +  or  —  .  Hence  z  has  four  values  —  the  four  roots 
of  equation  II.  In  equation  V  there  are  apparently  eight 
values  of  z,  but  four  of  them  are  ruled  out  by  the  relation 
2  Vw  Vv  VM?  =  —  G. 

From  the  above  we  see  that  the  roots  of  the  quartic  are 
expressed  in  terms  of  u,  v,  w.  The  values  of  the  latter  are 
given  in  terms  of  the  coefficients  of  the  quartic  and  the  three 
roots  x1}  x.2,  xs  of  the  cubic  IV.  To  solve  the  quartic  by  the 
present  method  we  must,  therefore,  first  solve  the  reducing 
cubic.  There  are  many  other  algebraic  solutions  of  the  general 
quartic,  but  every  one  of  them  calls  for  the  solution  of  an 
auxiliary  equation  of  the  third  degree.  These  cubics  are  called 
resolvents. 

Ex.  1.  Under  what  conditions  can  a  quartic  be  solved  algebraically 
without  the  extraction  of  cube  roots  ? 

It  is  only  necessary  that  the  reducing  cubic  have  a  rational  root,  so  that 
the  other  two  roots  can  be  expressed  in  terms  of  square  roots.  Euler's 
cubic  answers  equally  well. 


SOLUTION    OF   THE   CUBIC   AND    QUARTIC  73 

Ex.  2.  Show  that  the  reducing  cubic  of  x*  +  2  xa  +  x2  —  2  =  0  has  a 
rational  root.  Solve  the  quartic  by  square  roots. 

Ex.  3.  Show  that,  in  general,  the  values  of  x  and  y  in  x2  +  y  =  a, 
y*  +  x  =  b  cannot  be  found  algebraically  without  the  extraction  of  cube 
roots. 

Ex.  4.  Can  all  the  values  of  x  and  y  in  x2  +  y  =  11,  y2  +  x  =  7  be 
found  without  the  extraction  of  cube  roots?  For  solutions,  see  the 
Am.  Math.  Monthly,  Vol.  VI.,  p.  13,  Vol.  VII.,  p.  169 ;  see  also  Vol.  X., 
p.  192. 


.-V  . 


CHAPTER   VI 

SOLUTION  OF  BINOMIAL  EQUATIONS   AND   RECIPROCAL 
EQUATIONS 

63.   The  Binomial  Equation. 

xn  —  a  =  0, 

where  a  is  either  real  or  complex,  may  be  solved  trigonometri- 
cally  as  follows.  Let 

xn  —  a  =  r  I  cos  (2  for  +  0)  +  i  sin  (2  kw  +  0)  \ , 

where  k  may  assume  any  integral  value.  Then,  by  De  Mqivre's 
Theorem, 

2for  +  0   ,         .     2for  +  01 

s —      —  +  tsm—         -v- 
n  n       J 

By  assigning  to  Jc  any  n  consecutive  integral  values  we  obtain 
n  values  for  x  and  no  more  than  n,  since  the  n  values  recur  in 
periods. 

It  is  readily  seen  that  the  roots  are  all  complex  when  a  is  a 

complex  number.     For,  to  obtain  a  real  root,  — must  be 

n 

zero  or  a  multiple  of  TT  ;  that  is,  2  kir  +  6  will  be  zero  or  a 
multiple  of  TT;  hence  a  itself  must  be  real,  which  is  contrary 
to  supposition. 

When  o  =  + 1, 

then  xn  =  1  =  cos  2  for  +  i  sin  2  for, 

2  for  2  for 

and  a;  =  cos 1- i  sin »  I 

n  u 

74 


BINOMIAL   AND   RECIPROCAL  EQUATIONS 


75 


where  Jc  may  be  assigned  the  values  0,  1,  •  ••,  (n  —  1).     If  n  is 
odd,  then  k  =  0  is  the  only  value  of  k  which  yields  a  real  root, 

viz.  x  =  ~L.     If  n  is  even,  then  only  the  values  Jc  =  0  and  k  =  - 
yield  real  roots,  viz.  x  =  1  and  x  =  —  1. 

When  a  =  —  1, 

then  a"  =  —  1  =  cos  (2  A;  +  I)TT  -f-  i  sin  (2  fc  +  I)TT, 

where  A;  may  take  the  values  0,  1,  •••,  (n  —  1). 


Then 


=  cos 


2  k  4- 1 

There  'can  be  no  real  roots,  unless  is  an  integer,  and 

n 

therefore  n  an  odd  number.     If  n  =  2  Jc  +  1,  that  is  k  =  n~    , 
we  obtain  the  real  root  x  =  —  1. 

-A  64.    Geometrical  Interpretation  of  the  Roots  of  x"  =  a.     The  n 

roots  may  be  represented  graphically  in  the  Wessel's  Diagram 
(§  22)  by  n  lines  drawn  from  the  centre  of  a  circle  of  radius  -\/r 
to  points  on  its  circumference  and 
dividing  the  perigon  at  the  centre 

o 

into  equal  angles  of  —  radians. 
n 

Thus,  let  n  =  3  and  r  =  1.  The 
three  cube  roots  of  unity  are  seen 
from  I,  §  63,  to  be  1,  -  \  +  \  V^3, 
—  |-  — -i-V— 3.  They  are  repre- 
sented, respectively,  by  the  lines 
OA,  OB,  00.  These  lines  make 
with  each  other  angles  of  f  ir 
radians  or  120°.  The  circumference  is  divided  into  three  equal 
parts.  In  the  general  case  the  circumference  is  divided  into 
n  equal  parts.  Hence  the  theory  of  the  roots  of  unity1  is 
closely  allied  with  the  problem  of  inscribing  regular  polygons 
in  a  circle  or  the  theory  of  the  Division  of  the  Circle.  This 


76  THEORY   OF   EQUATIONS 

subject  has  been  worked  out  mainly  by  C.  F.  Gauss,  1801,  and 
will  be  treated  more  fully  in  Chapter  XVII  under  the  head 
of  Cyclotomic  Equations. 

65.  Roots  of  Unity.  We  give  a  few  general  properties  of 
the  ?ith  roots  of  unity,  some  of  which  are  evident  from  pre- 
vious considerations. 

I.  Tlie  equation  xn  =  1  has  no  multifile  roots. 

Here  f(x)  =  xn  —  1,  f'(x)  =  nxn~l.  Since  f(x)  and  f'(x)  have 
no  common  factor  involving  x,  there  are  no  multiple  roots 
(§  21). 

II.  If  a  is  a  root  of  xn  —  1  =  0,  then  a*  is  also  a  root,  k  being 
any  integer. 

Since  a"  =  1,  it  follows  that  «"*  =  1  or  (a*)n  =  1,  where  k  is 
zero  or  any  integer,  positive  or  negative.  Hence  «*  is  a  root 
of  unity.  As  there  are  only  n  roots,  it  is  evident  that  the 
powers  of  a  are  not  all  distinct  from  each  other,  and  «*  is  a 
periodic  function. 

III.  Ifm  and  n  are  prime  to  each  other,  the  equations  of  —  1  =  0 
and  xn  —  1  =  0  have  no  common  root  except  1. 

First  we  prove  the  theorem  :  If  m  and  n  are  prime  to  each 
other,  then  it  is  always  possible  to  find  integers  a  and  b  such  that 

mb  —  na  =  ±  1.     The  fraction  —  may  be  expanded  into  a  ter- 

n 

minatiug  continued  fraction,  say 

m  _  1 

n  ,  1 

q  ~\ — 

r 

•  77O  ~4~  1     f)(OT    I    1  ^    |j  o* 

The  successive  convergents  are  p,  ^-J — ,  ^w   ^ — i— L —     Sub- 

q          qr  + 1 

tracting  the  last  but  one  convergent  from  the  last,  we  obtain 
a  fraction  whose  numerator,  jxj(<]r  +  !)+(/»•  —  (pq  +  l)(^r  +  1), 
is  seen  to  be  equal  to  —  1.  (By  mathematical  induction  it  may 


BINOMIAL  AND    RECIPROCAL   EQUATIONS  77 

be  shown  that  if  —2=1  and  —  -  are  any  two  successive  conver- 

v,_i  vn 

gents,  then  wnVn-i  —  wn_iVn  =  ±  1.)     But 


m  =p(qr  +  1)  +  r,  n  =  qr  +  1  ; 
hence,  if  we  take  a=pq-\-l,  b  =  q,  we  have 

m&  —  an  =  ±  1.  Q.E.D. 

Now,  if  possible,  let  a  be  a  root  common  to  xm  —  1  =  0  and 
of  —  1  =  0.  Then  am  =  1,  «"  =  1  and  «mi  =  1,  «na  =  1,  where  a 
and  6  are  numbers  which  satisfy  the  relation  mb  —  na  =  ±  1. 
Hence,  amb-n°  =  l,  a±1  =  l,  or  «  =  1.  That  is,  1  is  the  only 
root  common  to  the  two  equations. 

jlr  IV.  If  h  is  the  highest  common  factor  of  m  and  n,  then  roots 
of  &  —  1  =  0  are  common  roots  ofxm  —  l  =  Q  and  xn  —  1  "=  0. 

We  have  m  =  hm',  n  =  hn',  where  m'  and  n'  are  prime  to  each 
other.  Hence  it  is  possible  to  find  integers  a  and  b,  such  that 
m'b  —  n'a  =  ±  1.  Multiplying  by  h,  we  get  mb  —  na  =  ±  h. 

Now,  if  a  is  a  common  root,  we  have  am  =  !,«"  =  1,  amb~na  =  1, 
or  «±A  =  1.  This  means  that  a  is  a  root  of  x"  —  1  =  0. 

V.  If  a  is  a  complex  root  of  aP  —  1  =  0,  n  being  prime,  then 
the  roots  are  1,  a,  a2,  «3,  •••,  a""1. 

By  II,  1,  a,  a2,  ••-,  a""1,  are  all  roots  of  the  equation.  They 
are  all  different  ;  for  suppose  ap  =  «*,  then  cf~q  =  1.  But  by 
III,  a;"  —  1  =  0  and  xp~q  —  1  =  0  cannot  have  a  root  in  common, 
since  n  and  (p  —  q)  are  prime  to  each  other.  Hence  the  equa- 
tion ap~q  =  1  is  impossible,  and  all  the  roots  are  included  in 
the  series  !,«,•••,  a""1. 

VI.  The  roots  of  the  equations 

xp  —  I  =  0,  '&  -  1  =  0,  of  -  1  =  0,  -. 
all  satisfy  the  equation  x™'"'  —  1  =  0. 

For  if  a  is  a  root  of  xp  —  1  =  0,  then  a?  =  1  and  (ap)fr*"  =  1, 
or  a?*""  =  1.  That  is,  «  is  a  root  of  xP"""  —  1  =  0. 


78  THEORY    OF    EQUATIONS 


66.  Primitive  Roots  of  Unity.  A  root  of  xn  —  1  =  0  is  called 
a  primitive  root  of  that  equation,  if  it  is  not  at  the  same  time  a 
root  of  unity  of  lower  degree. 

Take  of  -  1  =  0.  By  VI,  §  65,  the  roots  of  x2  -  I  =  0  and 
or5  —  1  =  0  are  roots/)  f  a6  —  1  =  0.  These  common  roots  are  1, 
—  1,  —  -|-  ±  ^  V  —  3.  The  other  two  roots  are  found  by  solving 
3?  -f  1  =  0  ;  they  are  +  \  ±  \  V—  3,  and  are  seen  to  be  primitive 
roots  of  x6  -  1  =  0. 

I.  We  proceed  to  show  that  primitive  roots  of  unity  exist  for 
every  degree  n. 

If  n  is  prime,  then,  by  III,  §  65,  x"  —  1  =  0  has  no  root  in  com- 
mon with  a  similar  equation  of  lower  degree,  except  the  root  1. 
Hence  all  the  roots  of  xn  —  1  =  0,  except  the  root  1,  are  primi- 
tive roots. 

If  n=pm,  where  p  is  a  prime,  every  exact  divisor  of  pm,  ex- 
cept pm  itself,  is  an  exact  divisor  of  pm~l.  Hence,  by  VI,  §  65, 
every  nth  root  of  unity  which  is  at  the  same  time  a  root  of 
unity  of  lower  degree,  must  be  a  root  of  xptn~  —1  =  0.  Since 
pm~l  is  a  factor  of  pm,  it  follows,  moreover,  that  every  root  of 
yfm~{  —  1  =  0  is  a  root  of  xpTn  —  1  =  0.  Thus,  there  are  pm~l  roots 
which  are  not  primitive,  and  the  number  of  primitive  roots  is 


If  n=pm'q>,  where  p   and   q  are   prime,  then  there   are 
pm(  1  --  )  primitive  roots  of  xpm  —  1  =  0  and  q(  1  --  )  priini- 

\     p)  \     q) 

tive  roots  of  &'  —  1  =  0.  Now,  if  n  and  /?  are  two  primitive  roots 
of  these  equations,  respectively,  then  aft  is  a  primitive  root 
of  ccn  —  1=0.  For  suppose  («/?)''=!,  where  r<n,  then  «r=/9~r. 
By  II,  §  65,  ar  is  a  root  of  x»m  —  1  =  0  and  /3~r  is  a  root  of 
yfl'  —  1  =  0.  But  the  two  equations  can  have  no  root  in  common, 
except  unity,  since  pm  and  (f  are  prime  to  each  other,  by  III, 
§  65.  Hence  r  cannot  be  less  than  n.  Since,  by  II,  §  65,  «"=! 


BINOMIAL   AND    RECIPROCAL   EQUATIONS  79 

and  ft"  =  1,  it  follows  that  («/8)n  =  1,  and  aft  is  a  primitive  root 
of  xn  —  1  =  0.     Since  there  are 


PJ    \       <2; 

such  products  a  •  ft,  this  expression  gives  also  the  number  of 
primitive  nth  roots  of  unity. 

It  is  easy  to  extend  this  proof  to  the  case  where  n=pmq'rt  •••. 

II.  We  give,  without  proof,  the  theorem  that  if  a  is  a  primi- 
tive nth  root  of  unity,  then  ar  is  a  primitive  nth  root  of  unity 
always  and  only  when  r  and  n  are  prime  to  each  other.     This 
theorem  enables  one  to  find  all  the  primitive  nth  roots  from  one 
of  them.* 

III.  The  roots  of  the  equation  xn  —  l  =  0,  where  n  =paqb  •••  re 
and  p,  q,  •••  r  are  the  prime  factors  of  n,  are  the  n  products  of  the 
form  fty  -••  8,  where  ft  is  a  root  ofxpa  =  1,  y  a  root  ofxq>'  =  l,  •••, 
8  is  a  root  of  of*  =  1. 

Let  a  =  fty  •'•  8. 

Here  ft  represents  any  one  of  pa  values;  similarly,  y,  •••,  8 
represent,  respectively,  qb,  •••,  i*  values.  From  this  it  may  be 
shown  that  a  has  n  values,  which  are  the  n  roots  of  xn  —  1  =  0. 

For,  in  the  first  place,  we  have  fpa  =  1,  y?5  =  1,  •••,  8K  =  1 ; 
hence,  also,  ft"  =  1,  yn  =  1,  •••,  8"  =  1,  and,  therefore,  nn  =  1. 

In  the  next  place,  we  show  that  the  n  values  of  a  are  distinct. 
If  possible,  let  two  values  of  «  be  equal,  say 


Since  not  all  the  roots  in  the  left  member  of  I  can  be  equal, 
respectively,  to  the  roots  in  the  right  member,  let  ft'  and  ft"  be 
distinct. 

*  For  the  proof,  see  Burnside  and  Panton,  Vol.  I,  1899,  p.  96.  We  have 
followed  the  exposition  of  the  subject  of  the  roots  of  unity  given  by  these 
authors. 


80  THEORY   OF   EQUATIONS 

From  I  we  get 

(py  ...  8')«*""*  =  03"y"  ...  8")**"-, 
and  (y'  .-  S')"6""*  =(y"...8")'*-|*  =  l. 

We  have  /J"6""*  =  /r«5""*. 

Since  /?'  and  /3"  are  distinct  roots  of  xpa  =  1,  they  are  equal  to 
two  different  powers  of  one  and  the  same  primitive  root  /3,  and 
we  may  write 

Pl-pm+m'l     p»  =  p*', 

where  m'  and  m  +  m'  are  each  less  than  pa.     We  get 

O(m+m')qb  •••  r<?  _    Om  'ji—re 
P  —  P  ) 

or  £»•*•••  *  =  1. 

Hence,  /3  is  a  root  of  both  xpa  =  1  and  af«*-rC  =  1,  and  also  of 
aj*  =  1,  where  s  is  the  highest  common  factor  of  p"  and  mg*  •••  7-c. 
(Theorem  IV,  §  65.)  But  we  have  s  ;>  m,  hence,  s<_p°.  Thus, 
^8  must  be  a  root  of  an  equation  of  lower  degree  than  pa.  Since 
/8  is  primitive,  this  cannot  be,  and  equation  I  is  impossible. 

IV.  The  roots  of  of  —  1  =  0,  where  p  is  prime,  can  be  found 
from  the  roots  of  equations  of  the  form  ocP  =  A. 

•Let  wl  be  any  root  of  xp  =  1,  iv.2  any  root  of  xp  =  «.'„  w3  any 
root  of  xp  =  w2,  and  so  on,  and  finally  iva  any  root  of  xp  =  wa-l. 
Then  the  product  a  =  wlwz  •••  wa  represents  pa  distinct  roots  of 

0^  =  1. 

For,  since  w-f  =  1,  w2p  =  w19  etc.,  we  obtain  successively  the 
relations, 


wap  = 

•  •  wa_2 


BINOMIAL   AND    RECIPROCAL   EQUATIONS  81 

V.  The  solution  of  xn  —  1=0,  where  n  is  any  composite  number, 
is  reduced  to  the  solution  of  binomial  equations  in  yuhich  n  is  a 
prime  number. 

This  important  result,  of  which  further  use  will  be  made  in 
a  later  chapter,  follows  readily  from  the  theorems  III  and  IV 
of  this  paragraph. 

67.  Depression  of  Reciprocal  Equations.  A  reciprocal  equation 
of  the  standard  form  (§  31)  can  always  be  depressed  to  one  of  half 
the  dimensions. 

Divide  both  sides  of  the  given  reciprocal  equation 
+  a1x2m~1  +  •  •  •  +  ajic  -f-  a0  =  0 


by  xm,  and  we  get,  on  collecting  in  pairs  the  terms  which  are 
equidistant  from  the  beginning  and  end, 


Assuming  y  =  x  +  -,  we  obtain 
x 


X 
and  generally 


By  substitution  in  the  above  equation  we  obtain  an  equa- 
tion of  the  mth  degree  in  y.  From  the  relation  x  +  -  =  y  we 
see  that  two  values  of  x  may  be  deduced  from  each  value  of  y. 


82  THEORY   OF    EQUATIONS 

Ex.  1.   Find  the  primitive  roots  of  x2  -  1  =  0,  x3  -  I  =  0,  x*  -  1  =  0. 

Ex.  2.   Find  the  roots  of  x5  -  1  =  0. 

Dividing  by  x  —  1,  we  get  x*  +  x3  +  x2  +  x  +  1  =  0. 

Dividing  this  reciprocal  equation  by  x2  and  taking  x  +  -  =  y,  we  obtain 


Solving  a;2  —  x?/  +  1  =  0,  we  arrive  at  the  following  four  roots  : 
zi  =  -  i(l  +  \/5  +  iVlO-2\/5),  x2  =  -  3(1  -  V5  -  fVlO  +  2V5), 
X3  =  -  i(l  -  Vo"  +  iVlO  +  2  V5),  x4  =  -  i(l  +  V5  -  iVlO  -  2V5). 

These  four  are  primitive  fifth  roots  of  unity.  The  other  root  is  1. 
Show  that  Xa  =  Xi2. 

Ex.  3.   Find  the  roots  of  x6  -  1  =  0. 

Ex.  4.    Find  the  roots  of  x7  -  1  =  0. 

Dividing  by  x  —  1,  we  get  a  reciprocal  equation  in  the  standard  form 
which  can  be  depressed  to  the  cubic  y3  +  y2  —  2  y  —  1  =  0. 

Writing  z  —  y  +  $,  we  have  z3  —  |  s  —  ^  =  0.  By  §  69  we  obtain  for 
y  three  values,  «,  «i,  02,  where 

a  =  -  J  +  J  V^S  +  84  v^~3"  +  }  ^28  -  84v^l. 
From  x2  —  x?/  +  1  =  0  we  get  the  six  values 


which,  together  with  unity,  are  the  seventh  roots  of  unity. 
Ex.  5.    Find  the  roots  of  x8  —  1.     Which  are  primitive  roots  ? 

Ex.  6.    Find  the  roots  of  x9—  1  =  0. 

Extracting  the  cube  root,  we  get  x3  =  1  or  w  or  in-  and  x  =  1 ,  w,  w2, 
Vw,  wVw,  to2Vio,  VW2,  wVw2,  w'2Viif2,  where  w  and  w-  are  the  primi- 
tive cube  roots  of  unity.  Give  the  primitive  roots  of  x9  —  1  =  0. 

Ex.  7.  Give  a  trigonometric  solution  of  x10  —  1  =  0  and  state  which 
roots  are  primitive. 

Ex.  8.   Find  the  primitive  roots  of  x12  —  1  =  0. 

Ex.  9.    How  many  primitive  roots  has  x180  —  1=0? 

Ex.  10.    Find  the  sum  of  the  primitive  roots  of  xu  —  1  =  0. 


BINOMIAL   AND   RECIPROCAL   EQUATUONS  83 

Ex.  11.   By  trigonometry  find  approximate  values  for  the  roots  of 
xn  -  1  =  0,  x13  -  1  =  0. 

Ex.  12.    From  the  primitive  roots  of  x8  —  1  =  0  and  x5  —  1  =  0  find 
the  primitive  roots  of  x15  —  1  =  0. 

Ex.  13.   Form  the  equation  whose  roots  are  the  primitive  roots  of 
x21  _  i  =  0. 

There  are  12  primitive  roots.     We  have 

x21  -  1  =  (x>  -  1)  (z14  +  x1  +  1). 

The  roots  of  x7  —  1  =  0  are  non-primitive  for  x21  —  1  =  0.  Since  xs  —  1 
is  a  factor  of  x21  —  1,  the  two  primitive  roots  of  x3  —  1  =  0  are  the  two 
remaining  non-primitive  roots  of  x21  —  1  =  0.  These  two  roots  are  roots 
of  x2  +  x  +  1  =  0.  Hence  (x14  +  x7  +  1)  -=-  (x/2  +  x  +  1)  =  0  is  the  re- 
quired equation.  This  is  a  reciprocal  equation  which  can  be  depressed 


Ex.  14.    If  —  V—  1  is  a  primitive  root  of  xn  —  1  =  0,  find  n.     If  —  V—  1 
is  a  non-primitive  root,  what  values  may  n  take  ? 


CHAPTER   VII 

SYMMETRIC  FUNCTIONS   OF  THE  ROOTS 

68.    Newton's  Formulae  for  Sums  of  Powers  of  Roots.      The 

sums  of  like  powers  of  the  roots  off(x)  =  Q  can  be  expressed 
rationally  in  terms  of  the  coefficients.  The  sum  of  the  pth 
powers  of  the  roots  a,  /3,  y,  8,  •••  of  the  equation  f(x)  =  0 
constitutes  a  symmetric  function  of  the  roots.  The  defini- 
tion and  elementary  discussion  of  symmetric  functions  were 
given  in  §  15.  Following  the  usual  notation,  we  designate 

2«p  by  s  ,  so  that  „ 

-  •- 


To  establish  Newton's  formulae,  write  (II,  §  20) 


firx\=  |  | 

x  —  a     x  —  ft     x  —  y 

The  indicated  divisions  can  be  exactly  performed,  §  3. 
If  /(«)  =  aj"  +  c^af-1  +  •  •  •  +  CWB  +  an, 


we  get  =  xn~l  +  (a  +  a^z"-2  -f  (a2  +  o^a  +  a2)z"-3  H  ---- 

a;  —  a 

+  (am  +  aia"-1  +  Oaa™-2  H  -----  f-  a,,)*"-*-1  -i  ----  . 

Similarly,  performing  the  divisionsof  J-&-}     J\  ).}  ...  }  and 
adding  all  the  quotients,  we  obtain  "  ^ 


f'(x)  —  naf1"1  +  (s1 

+  0m  +  <h*m-l  +  «2«m-2  H 


84 


SYMMETRIC    FUNCTIONS    OF    THE    ROOTS  85 

By  §  19,  we  know  that 
/'(a?)  =  no;"-1  +  (n  —  l~)a}xn~2  +  (n  —  2)a2an-3  4  ----  +  «„_!. 

Equating  coefficients  of  -the  same  power  of  x  in  the  two  expres- 
sions for  f'(x))  we  have 

sl  +  nor^  =  (n  —  !)<*!,  or  sl  +  al  =  0, 

s2  -f  aiSi"+  W(*2  =  (w  —  2)a2,  or  s2  +  a^  +  2  a2  =  0, 
and  generally,  when  ra<w, 

*„•+  «!«,_!  +  a2sm_2  H  -----  \-nam  =  (n  —  m)am, 
or  *„  +  Oi*m-i  +  ajSm-H  -----  h  am-A  +  mam  =  0.  I 

From  relations  I,  known  as  New  ton's  formulae,  we  derive  easily  : 

«!  =  —  Oj,         s2  =  a/  —  2  a2,         s3  =  —  af  +  3  a!a2  —  3  a3, 
s4  =  a^  —  4  a!2a2  +  4  a^  +  2  a22  —  4  a4, 

and  so  on,  up  to  sn_!.  To  extend  these  results  to  the  sums  of 
all  positive  integral  powers  of  the  roots,  viz.  sn,  sn+l,  •••,  multi- 
ply f(x)  =  0  by  xm~n,  where  m  >  n,  and  we  have 

xm  +  a^"1"1  +  a&m~~  +  •  •  •  +  anxm~n  =  0. 

Substituting  for  x  in  succession  the  roots  a,  /?,  y,  S,  •••  and 
adding  the  results,  we  get 

sm  +  aA,-i  +  a2s»-2  +  •••  +  «„««-„  =  0.  II 

If  we  give  m,  successively,  the  values  n,  n  +  1,  n  +  2f  •••  and 
observe  that  s0  =  n.,  we  obtain 

sn  +  arsn_!  +  aj8B_2  +  •••  +  nart  =  0, 


which  enable  us  to  find  expressions  for  sn,  sn+1,  •••. 

To  find  the  sum  of  negative  integral  powers  of  the  roots  of 
f(x)  =  0,  put  x  —  -  and  find  the  sums  of   the  corresponding 

y 

positive  powers  of  the  roots  of  the  transformed  equation. 


86 


THEORY   OF    EQUATIONS 


aj     1      0      0 

Oj        1       0 

ttj     1 
2a2     Ox 

,*3=- 

2a2     «!     1 

O  Gtg         Ctg         ^1 

,S<  = 

2  02     Oi     1      0 

O  Otg        CE-2        Clj         J. 
4  CZ^       (Z'Q       1*2       Cti 

The  values  of   sm  may  be  expressed  in  determinant  form 
as  follows : 


,  etc. 


69.    Coefficients  expressed  in  Terms  of  sm.     From  the  formulae 
of  §  68  one  readily  obtains  * 


,  etc. 


Ex.  1.  Find  the  sums  of  positive  powers  of  the  roots  of 

xt  +  x3  +  x2  +  x  +  1  =  0. 
We  have  si  =  —  ai  =  —  1, 

82  =  —  fliSi  —  2  az  =  —  1  > 

«s  =  —  OiS2  —  flzSi  —  3  as  =  —  1, 

84  =  —  aiSs  —  a2S2  —  cizSi  —  4  a*  =  —  1, 

and  so  on.     The  roots  are  the  primitive  fifth  roots  of  unity.     Verify  our 
result  for  s2  by  actually  squaring  the  roots  given  in  Ex.  2,  §  67. 

Ex.  2.   Find  the  sums  of  positive  and  negative  powers  of  the  roots  of 


.,    1 

_1 

Sl  1  0 
s2  Sj  2 

a-1 

«,     1     0    0 

s2    *!     2    0 

S3       Sj 

|3 

14 

3          2          1 

S3  S2  Si 

— 

^4        ^3        ^2        ^1 

«i  =  2,  s2  =  —  6,  83  =  —  10,  *4  =  +  18,  and  so  on.    To  get  s_m,  put  x  =  -, 

y 

and  the  equation  becomes  x3  —  f  x2  +  %  x  —  \  =  0.    Then  s_i  =  |,  $_2  =  /s, 
s_s  =  f  |,  and  so  on. 

Ex.  3.  Find  the  sums  of  positive  and  negative  powers  of  the  roots  of 


Ex.  4.   Show  that  if  the  sum  of  an  even  power  of  the  roots  is  zero  or 
negative,  the  equation  has  at  least  two  complex  roots. 

*  Ex.  5.   Show  that,  for  xn  —  1  =  0,  sm  =  n  or  0,  according  as  m  is 
divisible  or  not  divisible  by  n. 

Substitute  for  at,  •••,  an  their  values  in  I  and  II,  §  68. 


-'- 


SYMMETRIC    FUNCTIONS   OF    THE   BOOTS  87 

70.   Fundamental  Theorem  of   Symmetric  Functions.      Every 
rational  symmetric  function  of  the  roots  of  an  algebraic  equation 
\^can  be  expressed  rationally  in  terms  of  the  coefficients. 

To  begin  with,  we  shall  find  the  value  of  the  symmetric 
function  Sam/3p,  in  which  each  term  involves  two  of  the  roots. 

We  have 

*»  =  «"  +  /8"  +  /•  +  —, 


Multiplying,  we  get 

smsp  =  am+p  +  p**  4-  y 


that  is,  smsp  =  sm+p  +  2«m/5p, 

hence,  ^am^  =  smsp  -  sm+p.  I 

This  result  has  been  obtained  on  the  supposition  that  m  and  p  are 
unequal  integers.  If  they  are  equal,  then  the  terms  in  2lam(3p 
become  equal  two  and  two,  and  2«m/3p  =  2  2(«/3)m  =  sm2  —  s.2m. 
In  either  case  the  symmetric  function  is  expressed  as  a  rational 
function  of  the  sums  of  powers  of  the  roots.  But  by  §  68  the 
sums  of  like  powers,  sm,  can  be  expressed  rationally  in  terms  of 
the  coefficients  of  the  given  equation.  Hence  2«m/3p  can  be  ex- 
pressed rationally  in  terms  of  the  coefficients. 

Next  we  express  the  value  of  the  symmetric  function  2«m/?py7, 
where  each  term  involves  three  roots,  as  a  rational  function  of 
the  coefficients.  We  have 


ay 
s,  =  «»  +  /?« 

Multiplying,  we  have 


88  THEORY    OF    EQUATIONS 

The  terms  on  the  right-hand  side  constitute  three  sets,  rep- 
resented in  our  notation,  respectively,  by  2«m+('/3J>,  Sa1"/^*, 
2amj3y.  Hence 


Transposing  and  substituting  for  the  symmetric  functions  whose 
terms  involve  only  two  roots  their  values  as  determined  by  I, 
we  obtain 


s«,v«  -  *»+A  -  S™+«SP  ~  S™SP+<,  +  2  s>*+P+r      n 
This  supposes  that  ra,  p,  q  are  unequal.     If  m  =p,  we  have 

2  S(«0)"y  =  SA  -  ^  -  2  sm+ysm  +  2  s2m+g. 
If  m  =  p  =  5,  we  obtain  for  2am/?y  the  value  2  •  3  2(«/?y)n  and 
6  Sa»0V  =  sm3  -  3  s2msnl  +  2  ssm. 


Thus,  Sa^y*  may  always  be  expressed  rationally  in  terms  of 
the  coefficients  of  the  given  equation. 

This  method  may  be  continued  to  any  extent,  and  the  proof 
may  be  given  for  any  function  2«my8''y'8r  •  •  •. 

In  every  symmetric  function  thus  far  considered  all  the  terms 
were  of  the  same  degree;  the  function  was  homogeneous.  If 
any  rational  symmetric  integral  function  is  not-honiogeneous, 
thru  it  is  the  sum  of  two  or  more  homogeneous  symmetric 
integral  functions,  such  as  «  +  y3  +  y  +  aft  -f  «y  -f  /?y.  Hence 
it  is  cvidfiit  that  a  rational  symmetric  integral  function  can  be 
expressed  rationally  in  terms  of  the  coefficients,  whether  the 
function  be  homogeneous  or  not. 

Finally,  we  observe  that  no  fractional  function  can  be  sym- 
metric unless  it  can  be  so  reduced  that  its  numerator  and 
denominator  are  each  integral  symmetric  functions.  Hence, 
also,  a,  fractional  rational  symmetric  function  can  be  expressed 
rationally  in  terms  of  the  coefficients,  and  our  theorem  is 
established. 


SYMMETRIC    FUNCTIONS    OF   THE   ROOTS  89 

71.  By  the  aid  of  the  theorem  of  §  70  we  can  calculate  the 
value,  in  terms  of  the  coefficients,  of  any  rational  symmetric 
function.  But  this  method  is  laborious,  and  usually  other 
methods  are  preferable.  For  convenience  of  reference  we  state 
here  some  jof  the  results  obtained  in  §  15,  viz., 

For  the  cubic  Xs  +  ax2  +  bx  +  c  =  0, 
2a2/i?  =  3  c  —  ab, 


(«  +  0)08  +  y)(y  -f  «)  =  c  -  ab. 
For  the  quartic  x*  +  ax3  -+-  bx2  +  ex  +  d  =  0, 

2«2/3  =  3  c  -  ab, 
S«2j8*  =  &2  -  2  ac  +  2  d 

Ex.  1.     For  the  cubic  find   the  value,   expressed   in  terms  of  the 

coefficients,  of  W  +  *<*?. 

2«3/3  -H  Set2 

Ex/2J   For  the  quartic  find  the  value  of  the  irrational  symmetric 
function  V2«3/3. 

Ex.  3.     For  /(x)=0  calculate   S«i2a2c63,  where  «i,   «2,  —,  ctn  are 
the  roots. 

Multiply  S«i  =  —  «i 

\        'JJN  *"\. 

and  2cci«2«3  =  —  «§• 

In  the  product  the  term  (£1*0203  occurs  only  once,  the  term 
occurs  4  times.     Hence, 


and  Sai2a2«3  =  «i«s  —  4  «4- 


If  the  calculation  is  carried  on  by  §  70,  II,  we  have,  since  p—q=\  and 

«.»    _   O 

2  Sai2«2«3  =  «2*i2  -  2  sgSi  -  s22  +  2  s4. 

Substituting  for  sx,  s2,  s3,  s4  their  values,  §  68,  and  carrying  out  the 
indicated  operations,  we  get  the  same  answer. 


90  THEORY   OF   EQUATIONS 


f 


Ex.  4.  Show  that  for  the  general  equation  f(x)  =  0,  the  general  form, 
in  terms  of  the  coefficients,  obtained  for  2ai2«22  is  the  same  as  for  the 
quartic  equation. 

Ex.  5.  Calculate  S«18«2  for  f(x)=  0  and  from  the  result  derive  the 
special  value  it  assumes  for  the  cubic. 

Ex.  6.  Calculate  2a12a2'2«3  for  the  quintic  equation.  Is  the  result 
the  same  for  the  general  equation  ? 

Ex.  7.     Find  the  value  of  the  symmetric  function 

,a  -  0)2+  (0  -  7)2+  (7  -  «)2  for  the  cubic  b<#?  +  3  M2  +  3  62«  +  &a  =  0. 
Deduce  the  same  result  from  V,  §  35. 

Ex.-,  8.  By  aid  of  §  35  compute  the  value  of  («  -  0)2(«  -  7)2(0  -  7)2 
for  the  cubic  xs  +  x2  +  x  +  1  =  0.  What  relation  has  this  symmetric 
function  to  the  discriminant  of  the  cubic  ?  How  many  values  does  the 
function  («  —  0)(«  —  7)(/3  —  7)  assume  when  the  roots  are  interchanged  ? 
Why  is  this  function  not  symmetric  ? 

Ex.  9.    Show  that  for  the  quartic 

x*  +  aix3  +  a2x*  +  asx  +  a±  =  0, 

(«1«2  +  «3«4)  («1«3  +  «2«4)  («1«4  +  «2«s)  =  O32  +  «12<Z4  ~  4  fl^. 

Ex.  10.   Show  that  for  this  quartic 

(«j8  +  7*)(7«  +  £5)  +  (00  +  75)  (07  +  ««)  +  (07  +  a8)(7«  +  05) 
=  a\dz  —  4  #4. 

*  Ex.  11.   Form  the  cubic  equation  having  for  its  roots 

a/3  +  75,  «7  +  05,  07  +  ad. 

Ex.  12.  Show  how  the  general  quartic  may  be  solved  with  the  aid 
of  the  roots  of  the  cubic  in  Ex.  11  and  the  relation  afiyd  =  at. 

Ex.  13.  How  many  different  values  will  the  function  «0  +  75  assume, 
as  the  roots  are  interchanged  in  every  possible  way  ? 

*  Ex.  14.    Find  the  equation  whose  roots  are 


Let  the  required  equation  be 

&  +  aix5  +  a^x*  +  agx3  +  cttx2  +  a&x  +  a$  =  0. 


SYMMETRIC    FUNCTIONS   OF   THE   ROOTS  f  91 


We  have  a\  —  0,  and  therefore  «2  =  Sppi  =  —  \  Sp2  =  —  6.     Multiplying 
Sppi  by  Sp,  we  have  3  2ppip2  +  Sppi2  =  0,  hence 

08  =  -  SPP1P2  =  i  SPP12  =  -  i  V  =  -  10. 

Multiplying  Sppip2  by  Sp,  we  obtain 

4  Sppipzps  +  2ppip22  =  0  ;  Sppip22  =  Sp22  •  Sppi  —  Sp28  •  Sp  +  Sp2* 

=  Sp22-Spp!  +  Sp2*  =  —  48, 
hence  04  =  12. 
Similarly,  we  get 

5  Sppip2p3p4  +  Sppip2p32  =  0,  Sppip2p32  =  Sp32  •  Sppip2  —  Sp3s  •  Sppx 

-  Sp35  =  -  300, 

hence  as  =  —  60.     We  have  ae  =  17. 

*  Exf  15.  )  Find  the  value,  in  terms  of  the  cdemcients  of  the  cubic,  of 
(a  +  w<£i-t/w2a2)8  +  (a  +  w2«i  +  "«2)35  where  w  is  a  complex  cube  root 
of  unity. 

*  Ex.  16.    Show  that  for  the  quartic 

x4  +  4  bix3  +  6  62x2  +  4  63*  +  bt  =  0, 
the  following  relations  hold  : 

2«6ai  =  1536  6i*&2  -  2304  &!2&22  +  432  623  -  256  bfba  +  672  616263 

-  48  632  +  16  6!264  -  36  6264. 
=  256  6i363  -  288  6i6263  +  48  632  -  16  6i264  +  12  6264. 

96  bibtba  -  48  632  -  48  6ifl64  +  24  6264. 
l3  _  216  623  -  288  616363  +  48  632  +  48  6i264  -  18  6264. 

=  6  6264. 

Set2  =  16  &i2  -  12  62. 
Sa2«i«2  =  16  6163  -  4  64. 

*  Ex.  17i  Find  the  cubic  whose  roots  are 


*  Ex.  18.   Show  that,  for  the  quartic  x*  +  «ia;8  +  a2a;2  +  a3a;  +  a4  =  0, 
we  have 

(a  +  ai  -  «2  -  as)  (a  -  ai  -  a2  +  as)  (a  -  ai  +  a2  -  a8) 
=  -  Cai8  -  4  ai«2  +  8  a3). 


CHAPTER  VIII 

ELIMINATION 

72.   Resultants  or  Eliminants.     Let  us  determine  the  condi- 
tion that  the  two  equations 


F(x)  =  c<ftf  +  Cj  x  +  c2  =  0, 

shall  have  a  root  in  common.  Designate  the  roots  of  the 
second  equation  by  fa,  fa.  The  necessary  and  sufficient  con- 
dition that  fa  or  fa  shall  satisfy  the  equation  /(#)  =  0  is  that 
f(fa)  or  /(/82)  shall  vanish  ;  in  other  words,  that  the  product 
f(fa)  '/(/?2)  shall  be  zero.  Multiplying  together 


=  a0fa2  +  «,  fa  +  a2, 
we  get 


+  «oai(&&2  +  fa-  fa}  +  <*oa2(fa2  +  fa2)  +  a,2  fa  fa 
+  aia2(fa  +  fa)  -f  a22. 

Multiplying  by  c02  and  substituting  for  the  symmetric  func- 
tions of  fa  and  fa  their  values  in  terms  of  the  coefficients  of 
F(x)  =  0,  we  have 

—  O0o1c1c2  +  aoC^Cj2  —  2  a0a<p0c2  +  a!2c0c2  —  ajaaCoC!  +  a22c02- 


This  expression  is  called  the  eliminant  or  resultant.  Its  van- 
ishing is  the  condition  that  the  given  equations  shall  have  a 
root  in  common. 

If  from  n  equations  involving  n  —  1  variables  we  eliminate 
the  variables  and  obtain  an  equation  R  =  0  involving  only  the 

92 


ELIMINATION  93 

coefficients  of  the  eqiiations,  the  expression  R  is  called  the 
eliminant  or  resultant  of  the  given  equations. 

In  the  above  example  the  elimination  was  performed  with 
the  aid  of  symmetric  functions.  This  method  generalized  is 
as  follows  : 

73.  Elimination  by  Symmetric  Functions.  To  find  the  con- 
ditions that  the  two  equations 

/(«)  =  aox"  +  a^-1  +  a2xn~2  H  -----  \-  an  =  0, 
F(x)  =  COOT  +  da?"-1  +  c-jaf-2  -}  ----  +  cm  =  0, 


shall  have  a  common  root.  For  this  purpose  it  is  necessary 
and  sufficient  that  some  one  of  the  roots  &,  /32,  •••»$»  of  .F(a3)  =  0 
shall  satisfy  /(#)  =  0,  in  which  case  the  product 


must  vanish. 
We 


Multiplying  these  together,  we  obtain,  after  substituting  for 
the  symmetric  functions  of  /?j,  f32,  •••,  /Bm  which  occur  in  the 
product  their  values  in  terms  of  c0,cl5  •••,  cm,  and  after  clearing  of 
fractions,  ^  =  ..-/(ft)  ./(ft)  .../(W. 

Here  R  is  the  eliminant  and  is  a  rational  integral  function  of 
the  coefficients  of  f(x)  and  F(x).  Its  vanishing  is  the  condition 
that  the  two  given  equations  have  a  root  in  common.  The  degree 
of  the  resultant  in  the  coefficients  of  the  given  equations  is  in 
general  m  +  n. 

It  is  easy  to  see  that  we  obtain  the  same  eliminant  by  sub- 
stituting the  roots  a^ch,  •••,  «„  of  f(x)  =  0,  in  succession,  for  x 
in  the  polynomial  F(x).. 


94  THEORY   OF   EQUATIONS 

74.    Euler's  Method  of  Elimination.     If  /(a)  =  0  and  F(x)  =  0, 
as  defined  in  §  73,  have  a  root  «  in  common,  we  may  write 


where 


the  coefficients  Al}  •••,  A^  and  Ci,  •••,  Cn,  being  undetermined 
quantities. 

We  obtain  easily  the  identical  equation  of  the  (m  +  n  —  l)th 
degree  /(*)  •  ^(aj)  =  F(x)  -f^x). 

Performing  the  indicated  multiplications  and  equating  coef- 
ficients of  like  powers  of  x,  we  obtain  m  +  n  homogeneous 
equations.  Eliminating  the  undetermined  coefficients,  we  ob- 
tain the  required  resultant. 

Thus,  find  the  resultant  of 


a2  =  0,  c<p?  +  C&  +  c2  =  0. 
If  they  have  a  root  in  common,  we  obtain  the  identity 
C2)  (a<p<?  +  ctix  +  02)  =  (A,x  +  A2)  (c<jx?  +  c^x  +  c^) 


or  (dav  —  ^o^or5  + 

—  Afa  =  0. 


Equating  coefficients, 

=  0,  ' 
=  0, 
=  0, 
C^nz  —  A^c.2  =  0. 

In  order  that  the  four  homogeneous  equations  I  may  be  con- 
sistent with  each  other  it  is  necessary  that 


ELIMINATION 


95 


a0 


0 


0 

a0 


=  0. 


This  vanishing  determinant  is  the  resultant. 

[To  recall  the  reason  for  this,  observe  that  if  each  member  of 
the  four  equations  I  is  divided  by  A2,  we  have  really  only  three 

CCA 
unknown  quantities,  viz.  —i,  — ^,  — i-     If  their  values,  which 

•^•2     ^-^J      -^"*-2 

may  be  obtained  from  the  first  three  equations,  are  substituted 
in  the  fourth  equation,  then  we  obtain  a  relation  between  the 
coefficients  of  the  two  given  equations  which  is  the  same  as  that 
expressed  by  the  above  determinant.] 


75.  Sylvester's  Dialytic  Method  of  Elimination.  To  eliminate 
x  between  f(x)  =  0  and  F(x)  =  0,  equations  of  the  degrees  n 
and  m,  defined  as  in  §  73,  multiply  the  first  successively  by 
x°,  or1,  #V--,  xm~l,  and  the  second  successively  by  of,  xl,  x2,"-,  a;""1. 


We  obtain  thus  the 


equations 


f(x)  =  0,  xf(x)  =  0,  a»f(x)  =  0, 
F(x)  =  0,  xF(x)  =  0,  x*F(x)  =  0, 


=  0, 
=  0. 


The  highest  power  of  x  is  m  +  n  —  1.  If  f(x)  =  0  and 
F(x)  =  0  have  a  common  root,  it  will  satisfy  all  the  m  +  n 
equations.  If  the  different  powers  of  x,  viz.  x,  x2,  or3,  •••,  xm+n~l, 
be  taken  as  m  -f  n  —  1  unknown  quantities  satisfying  m+n 
linear  equations,  it  is  evident  that  a  relation  must  exist  between 
the  coefficients  of  the  equations..  This  condition  of  consistency 
is  the  vanishing  of  the  resultant.* 

*  The  above  proof  of  Sylvester's  method  is  the  one  usually  given. 
Attention  should  be  called  to  the  fact  that  it  is  not  shown  there  that  the 
different  powers  of  x  have  values  that  are  consistent. 


96 


THEORY    OF    EQUATIONS 


Thus,  to  find  the  resultant  of 

f(x)  = 
F(x)  = 


and 
we  have 


/(ic)  = 
xf(x)  =  a<f 
F(x)  = 
xF(x)  = 


+ 


=  0, 

+  c2  =  0, 

-f  a3  =  0, 

=  0, 

+  c2  =  0, 
=  0, 
=0. 


That  the  four  unknowns  x,  x2,  Xs,  x*,  may  satisfy  the  five 
equations,  it  is  necessary  that 


J?= 


0 
a0 
0 
0 

Co 


a» 
0 

Co 


a1 
a2 
c0 


«3 
Ci 

c2 
0 


fflg 
0 

C2 

0 
0 


=0. 


R  is  the  resultant. 


76.  Discriminant  of  f(x)  =  0.  It  was  proved  in  §  21  that  if 
f(x)  ==  0  has  a  multiple  root,  that  root  satisfies  /  '(x)  =  0.  The 
condition  that  f(x)  =  0  and/'(#)  =  0  have  a  root  in  common  is 
expressed  by  the  vanishing  of  their  resultant.  The  resultant 
of  f(x)  —  0  and  /'(»)  =  0  is  called  the  discriminant  of  f(x)  =  0. 
The  discriminant  of  an  equation  f(x)  =  0  may  be  otherwise 
defined  as  the  simplest  function  of  the  coefficients,  or  of  the  roots, 
whose  vanishing  signifies  that  the  equation  has  equal  roots. 

If  f(x)  =  0  and/'(#)  =  0  have  a  common  root,  this  root  will 
satisfy  also  nf(x)  —f'(x)  =  0.  Instead  of  finding  the  resultant 
of  f(x)  and  /'(#),  we  may  therefore  find  the  resultant  of 
nf(x)  —f'(x)  =  0  and  /'(#)  =  0.  The  latter  mode  of  procedure 
is  preferable,  because  it  gives  us  the  resultant  clear  of  an 
extraneous  factor. 


ELIMINATION  97 

The  discriminants  of  the  general  quadratic,  cubic,  and 
quartic  are,  respectively,  as  follows  : 

*--^__ 

Quadratic  disc.         =  —  (6j2  —  60&2) ; 
Cubic  disc.,  §  35,     =  -  — fi  (G2  +  4  H3} ; 

°0 

Quartic  disc.,  §  51,  =  — -(Is  — •  27  J2). 

77.  Discriminant  expressed  as  a  Symmetric  Function  of  the 
Roots.  Since  the  discriminant  of  the  equation  f(x)  =  0  vanishes 
always  when  at  least  two  roots  are  equal,  but  under  no  other 
conditions,  it  follows  that  «t  —  «2  must  be  a  factor  of  the  dis- 
criminant. For  if  «i  and  a.2  are  the  equal  roots,  «j  —  «2  is  the 
only  simple  factor  which  will  vanish  because  of  this  equality. 
But  an  interchange  of  any  two  roots,  say  %  and  a.2,  must  not 
alter  the  numerical  value  or  the  sign  of  the  discriminant,  since 
the  discriminant  is  a  constant  when  the  coefficients  of  the 
equation  are  constants.  Hence  the  lowest  positive  power  to 
which  the  factor  «j  —  «2  can  occur  in  the  discriminant  is  the 
second  power.  In  other  words,  («:  —  «2)2  is  a  factor  of  the 
discriminant. 

Since  this  reasoning  applies  to  any  two  roots  whatever, 
(«!  —  «3)2.is  a  factor ;  also  («!  —  «4)2;  and  so  OH. 

Hence  the  product 

is  a  factor  of  the  discriminant.  If  the  multiplications  indicated 
in  this  product  were  carried  out,  each  term  would  be  of  the 
n(n  —  l)th  degree  in  the  roots. 

The  resultant  of  /(a?)  =  0  and  /'(«)  =0  may  be  expressed  by 

§73  as  a  «.fi(a\.f'(a)...f'(a) 

where  al}  «2,  ••-,«„  are  the  roots  of  f(x)  =  0.  One  term  of  this 
product  is  (wa02)"(«iW2  •  ••  tt,,)""1;  the  degree  of  this  term  in  the  roots 


98  THEORY   OF   EQUATIONS 

is  n(n  —  1).  This  product  is  homogeneous,  for  if  in  any  other 
term,  say  (n  —  I)na0"a1"(«1«2  •••»«„)  "~2,  we  substitute  for  the  co- 
efficients their  equivalents  in  terms  of  the  roots,  by  the  relations 

of  §  13,  say  «j  -f  «„  -+-  •••  +  «„  for  --  -,  we  see  that  this  term 

a0 
likewise  is  of  the  degree  n(n  —  1)  in  the  roots.     Hence  the 

product  !!(«!  —  «2)2  is  of  the  same  degree  in  the  roots  as  the 
resultant  of  f(x)  =  0  and  /'(#)  =  0,  and,  therefore,  as  the  dis- 
criminant of  f(x)  =  0.  Consequently,  this  product  can  differ 
from  the  discriminant  by  a  numerical  factor  only. 

Ex.  1.  Show  that  the  resultant  of  x2  -  x  -  42  =  0  and  x2  +  4  x  -  11  =  0 
is  zero,  proving  that  the  left  members  of  the  equations  have  a  common 
factor. 

Ex.  2.   Find  the  resultant  of 
OQXS  +  ajX2+  dzx  +  «3  =  0  and  Coo;3  +  CiX2  +  c^x  +  c3  —  0  by  Euler's  method. 

Ex.  3.  For  what  value  of  a  will  the  two  equations  Xs  +  ax2  +  x  —  1  =  0 
and  x2  +  3a;  +  7=0  have  a  root  in  common  ? 


Ex.  4.    Using  Sylvester's  meth6d  of  elimination,  find  the  discriminant 
of  boX3+3  b&t+S  62*4-63  =  0. 

Ex.  5.   Find  the  discriminant  of  xn  —  I  =  0.     Has  the  equation  equal 
roots  ?    ' 

Ex.  6.    Find  the  discriminant  of  xn+l  —  xn  —  x  +  1  =  0. 


CHAPTER   IX 

THE  HOMOGRAPHIC   AND   THE   TSCHIRNHAUSEN 
.  TRANSFORMATIONS 

78.   Homographic  Transfonnation.     All  the  transformations 
of  equations  explained  in  §§  27-34  are  special  cases  of  the" 
homographic  transformation,  in  which  x  is  connected  with  the 
new  variable  y  by  the  relation 

iA 

XZ-K/* 

\'a;  +  / 

where  X,  X',  p,  fi'  are  constants!     Thus,  if  X  =  —  p.'=  1,  A'  =  p.  =  0, 
then  y  =  —  x,  as  in  §  28;  if  A  =  /i'  =  l  and  A'=0,  then  y  = 
as  in  §  32. 

By  solving  for  x  we  readily  get 


If  this  value  of  a;  is  substituted  in  a  given  equation  of  the 
nth  degree,  we  obtain  a  new  equation  of  the  nth  degree  in  y. 

If  a,  /?,  y,  .  .  .  are  the  roots  of  the  original  equation  and  «',  /?', 
y',  .  .  .  the  corresponding  roots  of  the  transformed  equation,  then 
we  have 


Subtracting,  we  get  «'  -  ft'  =       ~,~    /        We   obtain 

similar  expressions  for  a'  —  y',  8'  —  ft',  8'  —  y',  etc.     If  now  we 
take  any  four  roots  a,  y3,  y,  8  and  the  corresponding  roots  «',  /8'; 

99 


100  THEOKY    OF    EQUATIONS 

y',  8',  we  obtain  by  means  of  these  expressions  the  following 


The  geometrical  significance  of  each  of  these  fractions  becomes 
apparent,  if   taking    0  as   origin,    we   put   a  =  OC,  ft  =  OA, 

y=OB,8  =  OD.     Then 


AC     AD 

and  the  fraction  on  the  right-hand  side  is  equal  to  —  7-=--  -- 

BC     BD 

This  is  the  cross-ratio  (anharmonic  ratio)  of  the  points  C  and  D 
with  respect  to  the  points  A  and  B.  See  Ex.  10,  §  113. 

Similarly,  the  left-hand  fraction  expresses  the  cross-ratio  of 
points  C1  and  D'  with  respect  to  points  A'  and  B'.  Hence,  if 
the  roots  a,  ft,  y,  8  represent  distances  on  a  line,  measured  from 
an  origin  O,  then  the  cross-ratio  of  the  four  points  thus  deter- 
mined is  the  same  as  the  cross-ratio,  similarly  formed,  of  the 
points,  determined  in  the  same  manner  by  the  corresponding 
roots  a',  ft',  y',  8',  of  the  transformed  equation. 

Thus,  we  have  on  the  same  line  two  ranges  of  points, 
a,  ft,  y,  8,  •••  and  «',  ft',  y',  8',  •••  such  that  the  cross-ratio  of  any 
four  points  of  one  range  is  equal  to  the  cross-ratio  of  the 
corresponding  four  points  on  the  other.  Such  ranges  are  called 
homographic  ;  hence  the  name,  homographic  transformation. 
To  a  point  in  one  range  corresponds  one,  and  only  one,  point  in 
the  other.  In  other  words,  there  is  a  one-to-one  correspondence 
between  the  two  ranges  of  points.  The  homographic  trans- 
formation is  the  most  general  transformation  in  which  this 
correspondence  holds.  We  proceed  to  consider  transformations 
which  are  not  usually  homographic. 

<y^\   79.   The   Most  General   Transformation.       TJie    most   general 
^•'rational  algebraic  transformation  of  the  roots  of  an  equation 
(    f(x)  =  0  of  the  nth  degree  can  be  reduced  to  an  integral  trans- 
formation of  a  degree  not  higher  than  the  (n  —  l)th. 


THE   HOMOGRAPHIC   TRANSFORMATIONS  101 

Every  rational  function  of  a  root  am  can  be  expressed  in  the 
form  of  a  fraction  whose  numerator  and  denominator  are  each 
rational  integral  functions  of  the  root,  viz. 


Multiplying  both  numerator  and  denominator  of  by  the 

same  quantity,  we  may  write  ^"m-' 

•  7i(«m_i)  •  h(am+l)  -  -  •  ft(Q 


We  see  that  the  denominator  h(a^)  •  7i(«2)  •  ••  /*(«„)  is  a  sym- 
metric function  of  the  roots  a1}  «2,  •  •  • ,  «„  of  the  equation 
f(x)  =  0.  By  §  70  this  function  can  be'  expressed  rationally 
in  terms  of  the  coefficients.  Hence  «m  can  be  made  to  dis- 
appear from  the  denominator  of  the  fraction  representing  the 

value  of  — — .     In  other  words.  is  reduced  to  an  integral 

h(am}  h(am) 

function  of  am. 

Again^  the  numerator  of  this  fraction,  viz. 


is  a  symmetric  function  of  the  roots  «!,•••  «m_i,  «m+i>  ••• «»  of  the 

f(x) 
equation  -^   —  =  0.     Hence  it  can  be  expressed  as  a  rational 

^"m> 

function  of  the  coefficients  of  this  equation.  These  coefficients 
are  rational  integral  functions  of  am  and  the  coefficients  of 
f(x)  =  0,  as  may  be  seen  by  performing  the  indicated  division. 

Hence  and  also  2i^a2  can  be  expressed  as  an  integral 

rational  function  of  «m.    Let  the  integral  function  G(am)  =  .  ,  "  • 

If  G(am)  is  of  a  degree  higher  than  the  nth,  divide  G(x)  by 
f(x),  and  we  obtain 


102  THEORY    OP    EQUATIONS 

where  the  degree  of  the  function  H  (x)  does  not  exceed  n  —  1. 
Now  write  am  for  x.  Since  /(am)  =  0,  we  have  G  («„)  =  #(«„,), 
and  the  theorem  is  proved. 

80.  The  Tschirnhausen  Transformation.  The  most  general 
rational  algebraic  transformation  of  a  root  of  the  equation 
f(x)  =  0  can  therefore  be  represented  by  the  integral  functions 
of  the  (n  —  l)th  degree 


y  =(?!  +  d&  +  d3icr  -\  -----  1-  dnxn~l. 

This  is  known  as  the  Tschirnhausen  transformation. 

By  its  aid  Tschirnhausen  succeeded  in  reducing  the  general 
cubic  and  quartic  equations  to  the  form  of  binomial  equations. 
We  shall  do  this  for  the  cubic, 

Sox3  +  3  bjo?  +  3  b&  +  63  =  0. 

We  assume  y  =  d^  +  dgc  +  x2,  where  dl  and  d2  are  coefficients 
whose  values  must  be  determined. 

Let  the  roots  of  the  jjiven  equation  be  al}  «2,  «3,  and  the 
corresponding  roots  of  the  required  equation  y3  —  c  =  0  be 
/3,  <a(3,  o>2/3,  where  o>  and  o>2  are  the  complex  cube  roots  of  unity. 

Then 

ft  =  d1  +  d2«j  4- 

w£  =  dl 


Adding,  we  obtain  3  dl  +  c^Sj  +  .s2  =  0. 

Multiplying  the  second  equation  by  o>,  and  the  third  by  o>2,  and 
adding,   we    have     (ccj  -f  wa2  +  o)2a3)d2  +  «i2  +  w«22  +  wW  =  0. 
Whence 

-,    .  -,    .  «<,«»  4- 

s  =  =  --  = 


«!  +  0)02  +  0)J«3 

Since  CD  may  represent  either  one  of  the  two  complex  cube 
roots  of  unity,  there  are  two  possible  values  for  this  fraction. 


THE   HOMOGRAPHIC    TRANSFORMATIONS  103 

By  a  somewhat  laborious  operation,  these  values  may  be  shown 
to  be  roots  of  the  quadratic 

(&0&2  +  &lV  +  (&0&3  -  W»  +  (&1&3  -  V)  =  0. 

The  coefficients  of  this  quadratic  being  known,  we  can  find 
its  two  roots,  hence  also  the  required  values  of  dx  and  d.2. 
Then,  multiplying  together  the  members  of  equation  I,  and 
substituting  for  the  symmetric  functions  of  a19  «2,  «3  their  values, 
we  arrive  at  the  value  of  c  in  y3  —  c  =  0. 

After  reducing  the  cubic  and  quadratic  to  the  binomial  form, 
Tschirnhausen  hoped  to  be  able  to  transform  the  general 
quintic  to  the  form  y5  —  c  =  0.  Since  this  form  admits  of 
algebraic  solution,  he  hoped  to  find  the  much-sought-for  gen- 
eral algebraic  solution  of  the  quintic.  But  in  the  determina- 
tion of  the  coefficients  d1}  d.2,  ds,  d4,  d5,  unlooked-for  difficulties 
presented  themselves,  calling  for  the  solution  of  an  equation 
of  the  24th  degree.  While  the  Tschirnhausen  transformation 
is  worthless  for  the  general  solution  of  the  quintic,  it  enables 
one  to  remove  the  second,  third,  and  fourth  term  of  the  quintic 
and  of  equations  of  higher  degrees. 


Ex.  1.   Reduce  «2  +  ax  +  b  =  0  to  the  binomial  form  by  the  Tschirn- 
hausen transformation. 

Ex.  2.    Find  the  integral  transformation  of  a  degree  not  higher  than 

the  second,  which  is  equivalent  to  the  transformation  y  •=  z  +    •  for  the 
cubic  xs  +  x2  -f  x  +  2  =  0. 


Here  -ZAEL  =  x2  +  («2  +  l)z  +  («22  +  «2  +  1), 


«22  +  1         («12  +  1)  («22  +  1)  («32  +  1)         «12«2'W 

=  («22  +  «2  +  I)2  -  «22,  y  =  -  (x  +  I)2-    An*. 


- 

V 

—  '  ~~~ 


CHAPTER   X 

ON   SUBSTITUTIONS 

81.  Notation.  In  the  arrangement  or  permutation  of  four 
letters,  a^a./i^  let  each  letter  be  replaced  by  one  of  the  others  ; 
put,  for  instance,  a4  for  aly  as  for  a2,  «!  for  a3,  and  a2  for  a4,  then 
this  operation,  called  a  substitution,  may  be  designated  by  the 
notation 


where  each  letter  is  replaced  by  the  one  beneath,  or  by  the 
notation  (a^a/ig),  where  each  letter  is  replaced  by  the  one 
immediately  following,  the  last  letter,  a3,  being  replaced  by  the 
first,  ax.  We  shall  use  more  frequently  the  second  notation. 

Observe  that 


and  that 


Just  as  the  substitution  (a1a4a2a3),  effected  upon  the  arrange- 
ment a,a2a3a4,  gives  the  new  arrangement  a^a^,  so  when 
effected  upon  a^ajC^,  it  gives  cua^c^. 

We  shall  agree  that  in  a  substitution  a  letter  may  be  replaced 
by  itself,  but  that  no  two  letters  can  be  replaced  by  the  same 
letter.  Accordingly 


is  a  substitution,  but  (a1asa^z9a4)  is  not,  because  in  the  latter 
flt  and  as  are  both  replaced  by  as. 

104 


ON   SUBSTITUTIONS  105 

Ex.  1.   Show  that  (xyzw)  is  the  same  substitution  as  (wxyz). 
Ex.  2.    Show  that  (ai«2  •••«»)  is  equal  to 


that,  therefore,  the  same  substitution  may  be  represented  in  several  ways 
and  that  its  form  is  consequently  not  unique. 


82.  Product  of  Substitutions.  By  the  notation  (a^  •••«»), 
(bfiz-'-bn)  we  mean  that  the  substitution  (a^  •••«„)  is  per- 
formed first  ;  then,  upon  the  result  thus  obtained,  the  substitution 


is  performed.  We  call  the  two  substitutions,  placed  in  jux- 
taposition, their  product  in  the  given  sequence. 

If   the   product   (1  2  3)(4  5  3)    be  applied   to   the   digits 

12345.   taken   in   their   natural   order,   the   substitution 

7  I 

(1  2  3)  yields  the  arrangement  23145.  The  substitution 
(4  5  3)  applied  to  this  result  gives  the  arrangement  24153. 
But  this  last  arrangement  may  be  obtained  from  the  first  by 
the  substitution  (1245  3).  Hence  the  product  of  (123) 
and  (4  5  3)  is  equivalent  to  the  single  substitution  (1245  3). 

The  indicated  product  (1  2  3)  (4  5  3)  may  be  carried  out  conveniently 
as  follows  :  1  is  replaced  by  2  in  the  first  substitution,  and  2  is  not  re- 
placed in  the  second  substitution  ;  hence  1  is  replaced  by  2  in  the  prod- 
uct. Again,  2  is  replaced  by  3  in  the  first  substitution,  3  is  replaced  by 
4  in  the  second  substitution  ;  hence  2  is  replaced  by  4  in  the  product. 
Likewise,  4  is  replaced  by  5  in  the  second  substitution  and  also  in  the 
product  ;  5  is  replaced  by  3  in  the  second  substitution  and  in  the  product. 
Hence  the  result  of  the  multiplication  is  the  substitution  (1245  3). 

Ex.  1.    Show  that  (4  5  3)(1  2  3)  =  (1  2  3  4  5). 

Ex.  2.    Show  that  (abed}  (acde)  =  (abdce)  . 

83.  Commutative  and  Associative  Law.  Notice  that  the 
product  of  (1  2  3)  (4  5  3)  is  not  the  same  as  the  product  of 
(4  5  3)(1  2  3).  On  the  other  hand,  we  see  that  (1  2  3)  (4  5) 
=  (4  5)(1  2  3)  and  that  (xy)(zw)(xz)(yw)  =  (xz)(yw}(xy)(zw). 


106  THEORY   OF   EQUATIONS 

Hence  it  follows  that  in  the  multiplication  of  substitutions  the 
commutative  law  is  not,  in  general,  obeyed.  However,  we  shall 
find  that  the  associative  law  is  always  obeyed. 

Ex.  1.    Show  that  if  sa,  Sj,,  sc  are  substitutions, 

=  SaStSc. 


Assume  that  sa  replaces  an  element  p  by  q, 
that  sj  replaces  an  element  q  by  r, 
that  se  replaces  an  element  r  by  s, 
then  sas4  replaces  an  element  p  by  r," 
and  sbsc  replaces  an  element  q  by  s. 

Hence,  s0  sb  sc,  (sa  s&)sc,  s°(s6  sc)  each  replace  p  by  s. 

84.  Identical  Substitution.      A  substitution  which  replaces 
every  symbol  by  that  symbol  itself  is  an  identical  substitution. 

Example:  (    *  2  3  ),  which  may  also  be  written  (a1)(a2)(ag).    In 

\ttiCt  jj^S/ 

(az)  the  letter  a1?  is  at  the  same  time  the  first  and  the  last  letter, 
hence  it  is  replaced  by  itself.  As  the  identical  substitution 
plays  a  role  analogous  to  that  of  unity  in  the  product  of 
numbers,  it  is  usually  represented  by  1. 

-* 

85.  Inverse  Substitutions.     The  inverse  of  a  given  substitu- 

tion is  one  which  restores  the  original  arrangement,  so  that  a 
given  substitution  and  its  inverse  constitute  together  an  identi- 
cal substitution.  Thus,  the  inverse  of  the  substitution 


«=(  7?r3  "  '  M  is  the  substitution  (  Ol^3 
\blb2bs---bnj  Vftia2«3 

Let  the  inverse  of  the  substitution  s  be  designated  by  s~l. 
Then  the  inverse  of  s"1  is  s.  The  fact  that  any  substitution,  fol- 
lowed by  its  inverse,  gives  us  the  original  arrangement  may  be 
expressed  by  the  symbolism  0 .  c-i  _ 


We  have  also  s~l  •  s 

where  s°  signifies  an  identical  substitution,  i.e.  sn  =  1. 


ON    SUBSTITUTIONS  107 

The  repetition  of  a  substitution  s  or  s~l,  r  times,  is  denoted 
by  sr  or  s~r.  Hence  exponents  are  used  here  in  much  the 
same  way  as  are  integral  exponents  in  algebra. 

86.  Cyclic  Substitutions.  If  we  suppose  the  letters  of  the 
substitution  (a^  •••  an)  to  be  placed  in  the  given  order  on  the 


360° 
circumference  of  a  circle  at  equal  intervals  of  -  —  ,  the  given 

tt 

substitution  is  equivalent  to  a  positive  rotation  of  the  circle 

360° 
through  --  .     Hence  such  a  substitution  is  called  a  cycle,  or 

a  cyclic  substitution,  or  a  circular  substitution.  The  product 
(abc  ••-  d~)(xyz  •••  w)  is  called  a  substitution  of  two  cycles. 
Similarly  we  have  substitutions  of  three  or  more  cycles.  The 

substitution    to     ""  t)  Jfl    consists    of    the    two    cycles, 

\^o  4  o  7   1  A  ()J 

(1  3  5)  (2  4  7  6)  ;  for  1  is  replaced  by  3,  3  by  5,  5  by  I/  and 
we  have  one  cycle  ;  again,  2  is  replaced  by  4,  4  by  7,  7  by  6, 
6  by  2,  and  we  have  the  second  cycle. 

In  this  manner  any  substitution  can  be  resolved  into  cycles 
so  that  no  two  cycles  have  a  digit  in  common.  This  resolution 
can  be  effected  in  only  one  way. 

A  cycle  may  consist  of  a  single  element,  say  (5).     The  sub- 

stitution (  •    J  may  also  be  written   (1  3  4)  (2)  (5),  or 

Vo  ^41  o  I 

(1  3  4)2  5,  or  (1  3  4). 

Ex.  1.   Find  the  cycles  of  the  substitution  (      '    ,      }. 

\cdafgkhej 

Ex.  2.  Verify  the  relations  (ac&)  (a&c)  =  1  ,  (a&c)  (abc)  =  (ac6)  ,  (aft)  (ac> 
=  (a&c),  (&c)(ac&)  =  (ac),  (6c)(6c)  =  l,  (o6c)(a<*)  =  l. 

Ex.  3.  In  which  of  the  following  products  is  the  commutative  la\v 
obeyed:  (a&c)(ac),  (6c)(ac&),  (&ca)(&ac)  ? 

Ex.  4.    Write  the  inverse  of  (a&cde). 


108  THEORY   OF  EQUATIONS 

87.  Finite  Number  of  Distinct  Substitutions.  The  number  of 
distinct  substitutions  which  can  be  performed  upon  a  finite 
number  of  elements  a^  •••  an  is  finite,  for  the  number  of 
substitutions  cannot  exceed  the  number  of  permutations,  and 
this  is  known  to  be  finite.  Hence,  if  upon  a^  •••  an  we  per- 
form an  unlimited  series  of  substitutions  s,  s2,  s3,  s4,  •••,  the 
results  of  those  substitutions  cannot  all  be  distinct.  There 
will  be  certain  powers  of  s  which  give  the  same  result  as  does  s 
itself.  Let  m  +  1  be  the  lowest  power  of  this  kind,  then  sm+l  =  s. 
This  may  be  written  sm-s  =  s.  Hence 

smss~l  =  ss~l  =  s°  =  1, 
and  sm  =  1. 

We  call  m  the  order  of  the  substitution. 

•  The  order  of  a  substitution  is  the  least  power  of  the  substitu- 
tion which  is  equivalent  to  the  identical  substitution. 

/1234\,,        ,      /1234\     .      /1234\ 
s  =  [  ),  then  r-sal  } ,  s3  =  ( 

^234  I/  \3  4  1  2/         V4 1  2  3/ 

S4  = 


Hence  m  + 1  =  5,  m  =  4,  and  s4  =  <s°  =  1,  s6  =  s2,  and  generally, 
s*n+r  =  S*". 

This  substitution  s  is  cyclic.  It  is  evident  that  the  order  of  a 
cyclic  or  circular  substitution  is  equal  to  the  number  of  its  elements 
(digits). 

If  *=(123)(45),  then  s2=(132),  ss=(45),  s4  =  (123), 
s5  =  (1  3  2)  (4  5),  s6  =  1.  Hence  the  order  is  6. 

If  «j,  ?i2,  n3,  •••  denote  the  number  of  elements  in  the  successive 
cycles  of  a  substitution,  then  its  order  is  a  number  exactly 
divisible  by  each  of  the  numbers  n},  n2,  ns,  ••• ;  that  is,  its  order 
is  the  least  common  multiple  of  nlt  n2,  ns,  ••-. 

Ex.  1.  Show  by  actual  substitution  that  the  order  of  s  =  (1  2)  (3  4  5). 
(6789)  is  12  or  the  L.  C.  M.  of  2,  3,  4. 


ON   SUBSTITUTIONS  109 

88.  Theorem.  The  product  t~lst  may  be  conveniently  obtained 
from  the  substitutions  s  and  t  by  performing  upon  each  cycle  of  s 
the  substitution  t. 

Let  s  =  (o&c».)(aW  •••)•.  • 

and  ( 


Take  any  one  of  the  letters  a,  (3,  y,  •••  ,  a',  ft',  y',  •••,  say  ft. 
By  t'1,  ft  is  replaced  by  b  ;  by  s,  6  is  replaced  by  c  ;  by  t,  c  is 
replaced  by  y.  Hence  by  t~lst,  ft  is  replaced  by  y. 

Now,  if  by  £  we  substitute  ft  for  6  and  y  for  c  in  the  cycles 
of  s}  then,  instead  of  the  sequence  b  c,  we  have  in  s  the  sequence 
fty,  which  replaces  ft  by  y,  as  before.  As  this  consideration 
applies  not  to  ft  alone,  but  to  any  letter,  the  theorem  is 
established. 

In  the  operation  t~lst,  t  is  said  to  transform  s;  the  operation 
is  called  a  transformation. 

Ex.  1.  If  s  =  (1  2  3)  (4  5  6  7),  t  =  (5  7  2  3),  then  r1  =(327  5).  To 
illustrate  the  theorem  just  proved,  apply  r1  to  the  arrangement  1234567 
and  we  get  1724365.  To  this  result  apply  the  substitution  s,  and  we 
have  2435176.  To  this  last  arrangement  apply  t,  and  we  obtain 
finally  3  457/26. 

This  same  final  arrangement  is  obtained  more  easily,  if  in  place  of  per- 
forming the  three  substitutions,  we  perform  upon  the  arrangement 
1234567  only  one  substitution,  namely  s'=  (1  3  5)  (4  7  6  2).  Now  s'  is 
gotten  from  s  by  performing  upon  each  cycle  of  s  the  substitution  t. 

Ex.  2.  If  s  =  (1  2  3)  (4  567)  and  t  -  (2  4  3  7),  find  t~lst  by  theorem 
in  §  88. 

Ex.  3.  If  s  =  («5)(cc?),  t  =  (aftc),  determine  the  result  of  operating 
with  t-^st  upon  the  arrangement  abed. 

89.  Transpositions.  A  transposition  is  a  cyclic  substitu- 
tion containing  two  elements.  Thus,  (aft),  (be),  (1  2)  are 
transpositions. 

Ex.  1.  Show  that  the  square  of  any  transposition  is  the  identical 
substitution,  i.e.  1, 


110  THEORY    OF    EQUATIONS 

90.  Theorem.     A  substitution  may  be  expressed  as  the  produd 
of  transpositions  in  an  unlimited  number  of  ways. 

We  can  easily  verify  that 

(123...«)  =  (12)(13)  -(In), 
and  that    (1  2  3)(4  567)  •••  =  (1  2)(1  3)  (4  5)  (4  6) (4  7)  •-.. 

From  this  it  appears  that  every  substitution  can  be  expressed 
as  the  product  of  transpositions. 

The  number  of  ways  of  doing  this  is  unlimited,  for  between 
any  two  transpositions  just  found  we  may  interpolate  the  indi- 
cated square  of  any  transposition  without  modifying  the  sub- 
stitution; or  we  may  prefix  or  annex  the  square  of  any 
transposition,  and  we  may  continue  this  ad  libitum.  Thus, 

dbc  =  (<-'h)((ir)  =  (ca)(ca)(a&)(&c)(6e)(oc). 

91.  Theorem.     Tlie  number   of   transpositions   into   which   a 
substitution  is  resolvable  is  either  always  even  or  always  odd. 

The  effect  of  any  transposition,  say  («!«2)  upon  the  square 
root  of  the  discriminant,  V2J,  is  to  change  its  sign.  To  show 
this  write  (§  77) 


(«„  -i  — «»)• 

The  transposition  («!  «2)  alters  the  sign  of  the  factor  («j  —  «2) 
and  interchanges  the  remaining  factors  of  the  first  row  with 
the  factors  of  the  second  row.  The  factors  in  the  remaining 
rows  remain  unaltered.  Hence  the  sign  of  V-D  is  reversed  by 
a  single  transposition. 

Since  any  substitution  can  be  expressed  as  the  product  of 
transpositions,  the  effect  of  any  substitution  on  V/>  must  be 


ON   SUBSTITUTIONS  111 


either  to  al4£r  or  not  to  alter  its  sign.  If  the  sign  of 
remains  unchanged,  the  substitution  must  contain  an  even 
number  of  transpositions  ;  if  the  sign  of  V/>  is  changed,  the 
number  of  transpositions  must  be  odd.  Hence  no  substitution 
is  capable  of  being  expressed  both  by  an  even  and  by  an  odd 
number  of  transpositions. 

92.  Even  and  Odd  Substitutions.  A  substitution  expressible 
as  the  product  of  an  even  number  of  transpositions  is  called 
an  even  substitution  ;  one  expressible  by  an  odd  number  of  trans- 
positions is  called  an  odd  substitution.  Identical  substitutions 
are  classified  as  even. 

Ex.  1.   Are  the  following  substitutions  odd  or  even  ? 


/I 
VI 


2  34  56\      ,_/!  2  3\  /4  56  7 

3  2  5  6  4/'  \2  3  l)     4  6  7  5 


s"  =  (45  6)(1  7462  3),  *'"  =  (123  4)3. 

*  Ex.  2.  Show  that  any  substitution  transforms  an  even  substitution 
into  an  even  substitution.  See  §  88. 

93.  Theorem.  All  even  substitutions  can  be  expressed  as  the 
product  of  cyclic  substitutions  of  three  elements. 

If  two  transpositions  have  one  element  in  common,  we  have 
an  equality  like  the  following  : 

(1  2)(1  3)  =  (1  2  3). 

If  two  transpositions  have  no  element  in  common,  we  have 
the  following  relation  : 

(1  2)(3  4)  =  (1  3  4)(1  3  2). 

Thus,  since  any  two  pairs  of  transpositions  are  expressible 
in  terms  of  cyclic  substitutions  of  three  elements  each,  it  fol- 
lows that  any  even  substitution  can  be  thus  expressed. 

Ex.  1.  Express  the  even  substitution  (1  2  3  4)  (2  4  5  6)  as  the  prod- 
uct of  cyclic  substitutions  of  three  elements. 


CHAPTER   XI 

SUBSTITUTION-GROUPS 

94.  Example  of  a  Group.     The  substitutions 

1,  (1  2  3),  (1  3  2),  I 

are  distinct  and  possess  the  property  that  the  product  of  any 
two  of  them,  in  whichever  sequence  they  are  taken,  is  equal  to 
one  of  the  three.  Thus, 

(1  2  3)(1  3  2)  =  (1  3  2)(1  2  3)  =  1. 
1(1  2  3)  =  (1  2  3)1  =  (1  2  3). 
1(1  3  2)  =  (1  3  2)1  =  (1  3  2). 

Moreover,  the  square  of  any  substitution  gives  a  substitution 
in  the  set.  For,  (1  2  3)2  =  (1  3  2),  (1  3  2)*  =  (12  3),  I2  =  1. 
The  three  substitutions  I,  possessing  these  properties,  are  said 
to  form  a  group. 

95.  Definition  of  Substitution-group.     A  set  of  distinct  sub- 
stitutions, the  product  of  any  two  and  the  square  of  any  one  of 
which  belorigTolihe  set,  is  called  a  group  of  substitutions,  or  a 
substitution-group. 

When  using  the  term  group  we  shall  always  mean  a  substi- 
tution-group. 

The  substitutions  (1  2),  (1  3),  (1  2  3)  do  not  form  a  group ; 
for,  while  each  substitution  is  distinct  and  while  some  of  the 
products  yield  substitutions  in  the  set,  others  do  not.  Thus, 
(1  3)(1  2)  yields  (1  3  2),  which  does  not  belong  to  the  set. 

Ex.  1.  Prove  that  the  product  of  three  or  more  substitutions  of  a 
group  is  a  substitution  belonging  to  the  group. 

112 


SUBSTITUTION-GROUPS  113 

96.  Degree  and  Order  of  a  Group.  The  number  of  elements 
(letters  or  digits)  operated  on  by  the  substitutions  of  a  group  is 
called  the  degree  of  the  group.  The  number  of  substitutions  in 
a  group  is  called  the  order  of  a  group.  Thus,  the  group 

1,  (abc),  (acb),  (ab),  (ac),  (be) 

involves  the  three  elements  a,  b,  c  and  has  six  substitutions. 
Hence  it  is  of  the  third  degree  and  sixth  order. 


Ex.  1.    Tell  the  degree  and  order  of  the  group  1,  ( 
Ex.  2.    Prove  that  the  identical  substitution  satisfies  the  conditions  of 
a  group. 

Ex.  3.  Show  that  any  positive  integral  power  of  a  substitution  of  a 
group  is  a  substitution  of  that  group. 

Ex.  4.    Prove  that  the  identical  substitution  belongs  to  every  group. 

*Ex.  5.  Prove  that  the  inverse  of  any  substitution  in  a  group  belongs  to 
the  group. 

Ex.  6.  Every  substitution  s  in  a  group  is  equal  to  the  product  of  two 
substitutions  of  the  group. 

97.  Theorem.  Upon  the  distinct  letters  av  a2  •••  an  there  can  be 
performed  n  !  substitutions  which  form  a  group. 

From  elementary  algebra  we  know  that  the  total  number  of 
permutations  of  n  distinct  letters,  taken  all  at  a  time,  is 

w(n  —  l)(n  —  2)  —3'2-l=n!. 

Take  any  one  permutation  P.  We  may  change  it  into  any  one 
of  the  other  permutations  by  performing  a  substitution.  But 
for  no  two  of  these  other  n  !  —  1  permutations  is  the  substitution 
the  same.  Hence  there  must  be  one  less  than  n  !  such  substitu- 
tions. Counting  in  the  identical  substitution,  we  have  in  all  n  ! 
substitutions. 

These  n  !  substitutions  form  a  group.  For  with  any  one  of 
them  operate  upon  the  permutation  P,  then  upon  the  result 
thus  obtained  operate  with  the  same  or  any  other  substitution. 
The  second  result  will,  of  course,  be  some  one  of  the  n  !  permu- 


114  THEORY   OF   EQUATIONS 

tations  which  can  be  obtained  from  the  permutation  P  directly 
by  performing  one  of  the  given  substitutions.  Thus  it  follows 
that  the  product  of  any  two  substitutions  or  the  square  of  any 
substitution  is  equivalent  to  one  of  the  given  substitutions. 

Ex.  1.  The  letters  a\a<i<is  admit  of  the  six  permutations,  aia2«3,  ai«3a2, 
«2#i#3,  «2«s«i,  as«iffl25  Qsctzfti-  Show  that  these  six  permutations  are 
obtained,  respectively,  from  a\a^as  by  performing  the  substitutions  1, 
(«i)(a2«3),  (ai«2)(«3),  («i«2«3)5  (ffliasag),  (ai«3)(«2)-  Show  that  these 
substitutions  form  a  group. 

98.  Symmetric  Functions  and  Symmetric  Group.    A  symmetric 
function  of  n  letters  a1?  a2,  •  •  •,  an,  being  unaltered  in  value  when 
any  two  of  the  letters  are  interchanged,  undergoes  no  change 
in  value  when  it  is  operated  on  by  a  substitution  belonging  to 
the  group  given  in  the  preceding  theorem.     Because  of  this 
in  variance  the  symmetric  function  is  said  to  belong  to  that  group, 
and  the  group  bears  the  name  of  symmetric  group. 

Ex.  1.  By  applying  each  of  the  substitutions  of  the  symmetric  group 
1,  (aidzctz),  (aiflso^),  (#2^3) .  («i«s)>  («i«2)»  show  the  invariance  of  the 
symmetric  function,  a\az  +  a\a$  +  a2az- 

99.  Theorem.     All  even  substitutions  of  n  letters  form  together 
a  group. 

Even  substitutions  are  each  resolvable  into  the  product  of  an 
even  number  of  transpositions,  §  92.  Hence  the  product  of  any 
two  of  them  and  the  square  of  any  one  of  them  yield  even 
substitutions. 

Ex.  1.  With  the  letters  a,  6,  c  we  can  form  three  transpositions  (a&), 
(ac),  (6c).  Taking  the  products  of  every  two  of  these  in  either  sequence 
and  the  square  of  every  transposition,  we  obtain  the  following  distinct 
substitutions,  all  even,  which  form  a  group: 

1,  (a&c),  (rtcfe). 

Ex.  2.  Show  that  the  odd  substitutions  of  n  letters  do  not  form  a 
group. 


SUBSTITUTION-GROUPS  115 

100.  Alternating  Functions  and  Alternating  Groups.  Let 
ai>  Q-Z)  •••>  a»  be  n  magnitudes,  all  different.  A  function  of 
these,  such  that  an  interchange  of  any  two  of  them  changes 
the  sign  of  the  function,  is  called  an  alternating  function. 

Example :  (a^  —  a2)(a1  —  a3)(a1  —  a4)  •  •  •  (o^  —  an) 


(an_!  -  an). 

An  even  substitution  performed  upon  this  function  will  not 
alter  its  value.  For,  an  even  substitution,  which  consists  of  an 
even  number  of  transpositions,  will  reverse  the  sign  of  the 
function  an  even  number  of  times,  and  will,  therefore,  restore 
the  function  to  the  original  sign. 

Since  the  even  substitutions  of  n  letters  leave  an  alternating 
function  unaltered  in  value  while  all  the  odd  substitutions 
reverse  its  sign,  the  group  comprising  all  these  even  substitu- 
tions is  called  the  alternating  group  of  the  nth  degree.  Because 
of  this  in  variance  for  all  the  even  substitutions,  but  for  no 
others,  the  alternating  function  is  said  to  belong  to  the  alternat- 
ing group. 

*  Ex.  1.  Show  that  the  square  root  of  the  discriminant  of  an  equation 
of  the  nth  degree,  expressed  as  a  function  of  the  roots,  is  a  function  which 
belongs  to  the  alternating  group  of  the  nth  degree. 

101.  Cyclic  Functions  and  Cyclic  Groups.  The  powers  of  any 
substitution  form  a  group.  The  number  of  distinct  substitutions 
s,  sP,  s3,  •••,  resulting  from  taking  the  different  powers  of  the 
substitution  s,  cannot  exceed  the  order  of  the  substitution  (§  87). 
If  this  order  is  ra,  then  sm  =  l.  If,  therefore,  we  square  any  one 
of  the  ra  distinct  substitutions,  or  multiply  any  two  of  them 
together,  the  result  is  always  one  of  the  ra  distinct  substitu- 
tions. Hence  the  ra  distinct  substitutions  s,  s2,  s3,  •••,  sm  are  a 
group. 


116  THEORY   OF    EQUATIONS 

The  powers  of  the  cyclic  substitution  of  n  letters  (a^  •••  an) 
constitute  the  cyclic  group  of  the  degree  n. 

A  function  of  n  letters  which  is  unchanged  in  value  by  all 
the  substitutions  of  the  cyclic  group,  but  by  no  others,  is  called  a 
cyclic  function.  The  simplest  cyclic  function  belonging  to  the 
cyclic  group  of  the  degree  n  is 

c^Os*  +  a.2a/  -\ 1-  an_!an2  +  anaf. 

Ex.  1.  Show  that  the  function  ai«22  +  «2«32  +  «3«i2  belongs  to  the 
cyclic  group  1,  (010203),  (010302). 

Ex.  2.  Show  that  (01  +  a^  +  03  w2)3  belongs  to  the  cyclic  group  of 
degree  3,  w  being  a  complex  cube  root  of  unity. 

Ex.  3.  By  raising  (01030304)  to  powers  find  the  cyclic  group  of  the 
degree  4. 

102.  Transitive  and  Intransitive  Groups.     In  the  group 

1,  (1  2)(3  4),  (1  3)(2  4),  (1  4)(2  3) 

the  second  substitution  replaces  1  by  2,  the  third  replaces  1  by 
3,  the  fourth  replaces  1  by  4.  Similarly,  by  means  of  these 
substitutions  the  digits  2,  3,  or  4  can  be  changed  into  every 
other  digit  operated  on  by  the  substitutions  in  the  group.  This 
group  is  said  to  be  transitive. 

A  substitution  group  is  called  transitive  when  it  permits  any 
element  to  be  replaced  by  every  other. 

A  group  that  is  not  transitive  is  called  intransitive.  As  an 
example  of  the  latter  we  give  the  following  group, 

1,  (1  3),  (2  4),  (1  3)(2  4). 
Here  neither  1  nor  3  can  ever  be  replaced  by  either  2  or  4. 

103.  Primitive  and  Imprimitive  Groups.     If  in  the  transitive 
group  consisting  of  the  six  substitutions 

1,  (1  2  3  4  5  6),  (1  3  5)(2  4  6),  (1  4) (2  6) (3  6),  (1  5  3) (2  6  4), 
(165432) 


SUBSTITUTION-GROUPS  117 

the  digits  are  divided  into  the  two  sets  1,  3,  5  and  2,  4,  6,  then 
we  notice  that  each  of  the  three  substitutions  (123456), 
(1  4)(25)(36),  and  (1  6  5  4  3  2)  replaces  the  digits  of  one  set 
by  the  digits  of  the  other  set,  while  each  of  the  two  substi- 
tutions (1  3  5) (2  4  6),  (1  5  3) (2  6  4)  simply  interchanges  the 
digits  of  one  set  among  themselves.  This  group  is  called 
imprimitive. 

A  transitive  group  is  called  imprimitive  when  its  elements  can 
be  divided  into  sets  of  an  equal  number  of  distinct  elements,  so 
that  every  substitution  either  replaces  all  the  elements  of  one 
set  by  all  the  'elements  of  another,  or  simply  interchanges 
the  elements  of  one  set  among  themselves.  Otherwise  it  is 
primitive.  Example  of  a  primitive  group  : 

1,  (1  2  3),  (1  3  2). 

There  are  three  imprimitive  groups  oi  degree  four,  twelve  of 
degree  six,  and  no  imprimitive  groups  of  degree  two,  three,  and 
five. 

Ex.  1.  Show  that  no  group  whose  degree  is  a  prime  number  can  be 
impriuiitive. 

104.    List  of  Groups  of  Degree  Two,  Three,  Four,  and  Five.     We 

give  here  a  list  of  the  groups  of  the  first  five  degrees,  omitting 
only  the  group  1.  By  Gq(p)  we  mean  a  group  of  the  degree^) 
and  order  q.  We  give  also  the  notation  for  groups  used  by 
Cayley  and  others.  In  their  notation  the  symmetric  group  of 
degree  four  is  designated  by  (abed)  all;  eye  means  "cyclic" 
substitution ;  pos  means  "  positive  "  or  even  substitution.  For 
a  list  of  all  groups  whose  degree  does  not  exceed  eight,  see 
Am.  Jour,  of  Math.,  Vol.  21  (1899),  p.  326.  In  the  list  of 
groups  of  degree  n,  we  give  only  those  which  actually  involve  n 
letters.  But  it  must  be  understood  that  any  group  involving  less 
than  n  letters  may  be  taken  as  an  intransitive  group  of  the  nth 
degree.  For  instance,  G2(2)  =  1,  (a&)  may  be  written  as  a  group 
of  the  third  degree,  thus :  1,  (ab)(c). 


118  THEORY   OF   EQUATIONS 


DEGREE  Two. 

G^  =  (aft)  all  =  1,  (06). 
DEGREE  THREE. 

06<»>  =  (oftc)  all  =  1,  (o6c),  (ac6),  (06),  (oc),  (6c). 

Gs(3)  =  (abc)  eye.  =  1,  (a&c),  (ac6). 
DEGREE  FOUR. 

GMW  =  (a&cd)  all  =  (abed)  pos.  +  (06),  (cd),  (aebd),  (adbc), 
(be),  (ad),  (acdb),  (abdc),  (ac),  (6d),  (abed),  (adcb). 
pos.  =  1,  (ab)(cd),   (ac)(6d),  (ad)(6c),   (a6c), 
(acd),  (6dc),  (ad6),  (ac6),  (6cd),  (a6d), 
(ode). 

1,  (ac)(6d),   (ac),   (6d),    (ab)(cd),   (ad)(6c), 

(abed),  (adcb). 

=  (a6cd)  eye.  =  1,  (ac)(bd),  (abed),  (adcb). 
=  (abcd)4  =  1,  (a6)(cd),  (ac)(M),  (ad)(6c). 
=  (06  •  cd)  =  1,  (a6)(cd),  (aft),  (cd), 
#.w  =  (ac  .  6d)  =  1,  (ac)(6d). 
DEGREE  FIVE. 

GE120(5)  =  (abcde)  all  =  (abode)  pos.  +  (abed),  (abdc),  (abce), 
(abec),  (abde),  (abed),  (aebd),  (acdb), 
(acbe),  (aceb),  (acde),  (aced),  (adbc), 
(adcb),  (adbe),  (adeb),  (adce),  (adec), 
(aebc),  (aecb),  (aebd),  (aedb),  (aecd), 
(aedc),  (bcde),  (bdce),  (bced),  (bdec), 
(becd),  (bedc),  (abc)(de),  (acb)(de), 
(abd)(ce),  (adb)(ce),  (abe)(cd),  (aeb)- 
(cd),  (acd)(be),  (adc)(be),  (ace)(bd), 
(aec)(bd),  (ade)(bc),  (aed)(bc),  (bed)- 
(ae),  (bdc)(ae),  (bce)(ad),  (bec)(ad), 
(bde)(ac),  (bed)(ac),  (cde)(ab),  (ced)- 
(ab),  (ab),  (ac),  (ad),  (ae),  (be),  (bd), 
(be),  (cd),  (ce),  (de). 


SUBSTITUTION-GROUPS  119 


.  =  l,  (abcde),  (abced),  (abdec),  (abdce), 
(abecd),  (abedc),  (acbde),  (acbed), 
(acdbe),  (acdeb),  (acebd),  (acedb), 
(adceb),  (adobe),  (adecb),  (adebc), 
(adbec),  (adbce),  (aebcd),  (aebdc), 
(aecbd),  (aecdb),  (aedcb),  (aedbc), 
(abc),  (acb),  (acd),  (ode),  (ode),  (aed), 
(abd),  (adb),  (abe),  (ae6),  (acc),-(aec), 
(bed),  (6dc),  (6de),  (6ed),  (6ce),  (6cc), 
(cde),  (ccd),  (a6)(cd),  (o6)(ce),  (a6)(de), 
(ac)(6cZ),  (ac)(6e),  (ac)(de),  (oe)(6d), 
(oe)(6c),  (ae)(cd),  (ad)(6c),  (od)(6e), 
(ad)(ce),  (6c)(de),  (6d)(ce),  (6e)(cd). 
=  1,  (abcde),  (acebd),  (adbec),  (aedcb), 
(bced),  (acbe),  (aecd),  (abdc),  (adeb), 
(bdec),  (adce),  (abed),  (aebc),  (acdb), 
(be)(cd),  (ae)(bd),  (ad)(bc),  (ac)(de), 
(ab)(ce). 

#12<5>  =  (a6c)  all  (de)  =  1,  (o6c),  (ac6),  (a6c)(de),  (ac6)(de), 

(a6)(de),  (ac)(dc),  (6c)(de),  (a6), 
(ac),  (6c),  (de). 

G^jo^  =  (abcde)i0  =  1,  (abcde),  (acebd),  (adbec),  (aedcb), 
(be)(cd),  (ae)(bd),  (ad)(bc),  (ac)(de), 
(ab)(ce). 

G/l=  \(abc)  all    (de)}   pos  =  1,    (a6c),    (ac6),    (ab)(de), 

(ac)(de),  (bc)(de). 

Of  II  =  (abc)  Gyc.  (de)  =  I,    (de),   (abc),    (abc)(de),    (acb), 

(acb)(de). 
Gs(5)  =  (abcde)  eye.  =  1,  (abcde),  (acebd),  (adbec),  (aedcb). 


120  THEORY    OF    EQUATIONS 

Ex.1.  Show  that  the  order  of  any  alternating  group  is  — ,  where  n  is 
the  degree  of  the  group. 

Ex.  2.  Tell  by  the  orders  of  the  groups  which  of  the  groups  of  the  first 
five  degrees  are  the  symmetric,  which  are  the  alternating  groups. 

Ex.  3.  By  inspection,  find  which  of  the  groups  of  the  degrees  two,  three, 
and  four  are  transitive,  intransitive,  primitive,  imprimitive. 

Ex.  4.  Show  that  the  imprimitive  group  in  §  103  may  have  its  elements 
divided  into  the  three  sets  1,4;  2,  5  ;  3,  6,  and  that  it  is  imprimitive  with 
respect  to  these  sets. 

Ex.  5.  Show  that,  of  the  groups  of  the  fifth  degree,  three  are  intransi- 
tive, viz.  018(6),  G6&I,  £6(5)II. 

*  Ex.  6.   Show  that  the  intransitive  group  G^WIII  is  obtained  by  multi- 
plying every  substitution  of  the  group  1 ,  (ad)  by  every  substitution  of  the 
group  1,  (cd~). 

*  Ex.  7.    Show  that  the  intransitive  group  G^II  is  obtained  by  multiply- 
ing the  substitutions  of  the  group  1,  (a&c),  (ac6)  by  the  substitutions  of 
the  group  1,  (de)  ;  that  (?6<5>I  is  the  product  of  the  group  1,  (a&c),  (acfe) 
and  the  group  1,  (a&)  (de)  ;  that  GIZ^  is  the  product  of  GV3)  and  the  group 
1,  (de). 

Ex.  8.  Show  that  a  group  of  the  third  degree  may  be  regarded  as  an 
intransitive  group  of  a  higher  degree. 

105.  Sub-groups.  The  alternating  group  of  degree  4  is 
(§  104) 

1,  (1  2)(3  4),  (1  3)(2  4),  (1  4)(2  3),  (1  2  3),  (1  3  2),  (1  3  4), 
(1  4  2),  (1  2  4),  (1  4  3),  (2  3  4),  (2  4  3). 

We  observe  that,  of  the  12  substitutions,  the  following  four 
make  up  a  smaller  group  of  their  own : 

1,  (1  2)(3  4),  (1  3)(2  4),  (1  4)(2  3). 

Thus  we  may  have  groups  within  groups.  If  from  the  sub- 
stitutions of  a  group  we  can  pick  a  set  which  form  a  group  all 
by  themselves,  this  second  group  is  called  a  sub-group  of  the  first. 
The  terms  group  and  sub-group  are  only  relative.  A  sub-group 
considered  by  itself  is  called  a  group,  and  a  group  may,  in  turn, 
be  a  sub-group  of  another  of  still  higher  order. 


SUBSTITUTION-GROUPS 
Ex.  1.   By  inspection,  find  sub-groups  of 


Ex.  2.   How  many  sub-groups  has  GUW  ?     See  §  104. 

Ex.  3.   How  many  sub-groups  has  G\zw  ? 

Ex.  4.    What  sub-groups  has  (abcde)i0  ?     (a&c)  all  (de)  ?     (abcde)  all  ? 


106.  Theorem.  The  order  of  a  sub-group  is  a  factor  of  the 
order  of  the  group  to  which  it  belongs. 

Let  the  substitutions  of  the  sub-group  be  s1}  s2,  ss,  •  ••,  sn, 
and  let  t  be  any  substitution  of  the  group  which  does  not  occur 
in  the  sub-group.  Then,  by  the  definition  of  a  group,  we 
know  that 


are  all  substitutions  belonging  to  the  group,  but  none  of  them 
belong  to  the  sub-group  5  for  suppose  s^t  =  sr)  then 


Since  s^1  is  a  substitution  of  the  sub-group  (see  Ex.  5,  §  96),  it 
follows  that  its  product  with  sr,  namely  t,  belongs  to  the  sub- 
group —  which  is  contrary  to  supposition. 

Moreover,  the  new  substitutions  in  I  are  all  distinct  ;  for  sup- 
pose s2£  =  sst,  then  it  would  follow  that  s2  =  s5. 

If  the  substitutions  in  I  do  not  exhaust  the  substitutions  in 
the  group  not  belonging  to  the  sub-group,  then  suppose  the 
substitution  tt  is  among  those  left  over.  Then 

Sjti,  S<fj}  Sgti,   •••,  Snti, 

are  distinct  substitutions  of  the  group  not  found  in  the  list 
su  S2i  •••>  sn  for  reasons  just  mentioned  ;  nor  are  they  found  in  I  ; 
for  suppose  sj  =  s^,  then  ^  =  Sa"1^  =  M>  which  is  some  sub- 
stitution in  I,  a  conclusion  contrary  to  the  assumption  concern- 
ing tv  Continuing  in  this  way,  the  substitutions  of  the  group 
are  divided  into  sets  of  n  substitutions  each.  As  the  number 


122  THEORY   OF   EQUATIONS 

of  substitutions  is  assumed  to  be  finite,  this  process  must  come 
to  an  end,  and  we  have  the  sets 


o    /  O    "f  O/  O"f" 

Sjl,         Scf,         SgC,  •••,      Snl, 


The  total  number  of  substitutions  in  the  group  is  therefore  n 
times  the  number  of  sets,  or  (m  +  2)n.  But  (m  +  2)n  is  the 
order  of  the  group,  and  n  the  order  of  the  sub-group.  Hence 
the  order  of  the  sub-group  is  a  factor  of  the  order  of  the  group. 

107.  Index  of  a  Sub-group.     If  n  is  the  order  of  a  group  G 

and  ra  the  order  of  a  sub-group  GI,  the  quotient  —  is  called  the 

m 

index  of  GI  under  G.     Thus  the  index  of  an  alternating  group 
under  the  symmetric  group  of  the  same  degree  is  n  I  -=-  —  =  2. 

6 

Ex.  1.  Give  the  index  of  every  group  of  the  fifth  degree  under  the 
symmetric  group. 

Ex.  2.  Show  that  a  group  whose  order  is  prime  can  have  no  sub-group 
(except  the  substitution  1). 

108.  Normal  Sub-groups.  —  If  G±  is  a  sub-group  of  G,  and  s 
any  substitution  of  G  which  does  not  occur  in  Glt  the  groups  GI 
and  s~lG±s  are  called^onjugate  sub-groups  of  G.     By  the  trans- 
formation s~lGiS,  we  meanthe  resultTobtained  by  subjecting 
every  substitution  St  of  the  sub-group  Gl  to  the  transformation 


eve 
*-'.< 


If  GI  and  s~lGiS  are  identical  to  each  other,  whatever  substi- 
tution s  is  of  G,  GI  is  called  a  normal  sub-group,  or  a  self-conju- 
gate sub-group,  or  an  invariant  sub-group  of  G. 

109.  Simple  Groups.  —  A  simple  group  is  one  which  has  no 
normal  sub-groups,  other  than  the  group  consisting  of  the 
identical  substitution. 


SUBSTITUTION-GROUPS  123 

It  can  be  shown  that  the  alternating  group  of  every  degree 
above  four  is  simple  (§  198).  It  is  readily  seen  that  all  groups 
whose  order  is  a  prime  number  are  simple.  There  are  only  six 
groups  whose  orders  are  not  prime  numbers  and  do  not  exceed 
1092,  which  are  simple,  viz.,  the  groups  of  the  orders  60,  168, 
360,  504,  660,  1092.  Those  of  order  60  and  360  are  alternating 
groups  of  the  degrees  five  and  six,  respectively. 

A  group  which  is  not  simple  is  called  composite. 

Ex.  1.    Find  the  groups  conjugate  to  (?2(4)  under  (?i2(4). 

If  we  transform  Si  =  (ac)(bd)  by  s  =  (a&c),  we  get  s~lsis  =  (a6)(cd). 
In  the  same  way  transforming  st  =  1,  we  get  1.  Hence  a  group  conjugate 
to  C?2(4)  is  1,  (a6)(cd).  We  obtain  the  same  conjugate  group  by  taking 
for  s  the  substitutions  (acd)  and  (adb}. 

The  transformation  of  s-lG^s,  where  s  =  (feac),  yields  the  conjugate 
sub-group  (ad)  (fee),  1.  The  same  result  is  obtained  if  we  take  s  =  (ac6), 
(bed),  (abd),  or  (adc). 

Taking  s=  (ac)(bd)  or  (ad)(6c),  the  conjugate  groups  obtained  are 
identical  with  GV4^  The  distinct  conjugate  sub-groups  of  (?2(4)  under  GI%(*) 
are,  therefore,  1?  (ac)(&d), 


We  see  that  £r2(4)  is  not  a  normal  sub-group  of  C?i2(4). 

Ex.  2.    Find  the  conjugate  groups  of  GZW  under  G^  I. 

Ex.  3.   Find  the  conjugate  groups  of  Gy5)  II  under  G\45). 

Ex.  4.    Find  the  conjugate  groups  of  6re(5)  I  under  (?i2(5). 

Ex.  5.    By  actual  trial  show  that  #3<3>  is  a  normal  sub-group  of 
that  Cr2(4)  is  a  normal  sub-group  of  G^4'  II  ;  that  G^  II  is  a  normal  sub- 
group of  GV4'  ;  that  G^  I  is  a  normal  sub-group  of  G»w. 

Ex.  6.    Show  that  every  group  has  identity  as  a  normal  sub-group. 

Ex.  7.  Prove  that  the  alternating  group  G(n\n\  is  a  normal  sub-group  of 
the  symmetric  group  #(">„!  See  Ex.  2,  §  92. 

Ex.  8.    Prove  that  a  cyclic  group  of  prime  degree  is  simple. 

Ex.  9.  -Prove  that  the  alternating  group  embraces  all  circular  sub- 
stitutions of  odd  order,  but  none  of  even  order. 

Ex.  10.  The  substitutions  common  to  two  groups  constitute  a  group 
by  themselves,  the  order  of  which  is  a  factor  of  the  orders  of  the  two 
given  groups. 


124  THEORY    OF   EQUATIONS 

110.    Normal  Sub-groups  of  Prime  Index.      Of  special  interest 
in  the  theory  of  equations  are  the  series  of  groups 
p    p          p    p  i 

*  U   *»          )  *•   -*<+!>         >  x 

so  related  to  each  other  that  each  group  Pi+1  is  a  normal  sub- 
group of  the  preceding  group  Pi}  the  index  of  Pi+l  under  P, 
being  a  prime  number.  Such  an  assemblage  of  groups  is  called 
a  principal  series  of  composition.  If  the  restriction  of  a  prime 
index  is  removed,  then  the  assemblage  is  called  simply  a  series 
of  composition. 

Ex.  1.  Show  that  a  principal  series  of  composition  is  (a)  for  groups  of 
the  third  degree,  (?6(3),  Gy3V  1>  (&)  f°r  groups  of  the  fourth  degree,  G&f-*), 
#uW,  &tw  II,  G,<4),  1. 

Ex.  2.  Show  that,  for  the  group  of  the  fifth  degree  6r2o(5),  a  principal 
series  of  composition  is  6r2o(5),  Gw(5),  &5(5),  1. 

Ex.  3.    Show  that  Q+W  II  is  a  normal  sub-group  of  £8(4),  <n2(4),  and 


111.  Functions  which  belong  to  a  Group.  When  GI  is  a  sub- 
group of  G,  a  rational  function  of  n  letters  a1}  a2,  •••,  an  is  said 
to  belong  to  Gl}  if  the  function  is  unaltered  in  value  by  the  sub- 
stitutions of  G1}  but  is  altered  by  all  other  substitutions  of  G* 

*  If  the  coefficients  of  f(x)  =0  are  independent  variables,  then  its  roots 
are  independent  of  each  other.  A  function  of  the  roots  must  therefore  be 
looked  upon  as  having  an  alteration  in  value  whenever  the  function  experi- 
ences an  alteration  inform.  In  other  words,  when  the  roots  are  independent 
of  each  other,  two  functions  of  these  roots  are  equal  to  each  other  only  when 
they  are  identically  equal.  In  the  present  chapter  the  roots  are  so  taken. 

When  the  coefficients  of  /(a:)=0  represent  particular  numerical  values, 
its  roots  are  fixed  values.  Two  functions  of  these  roots  may  be  numerically 
equal  to  each  other  even  when  they  have  different  forms.  Hence,  in  an  equation 
whose  coefficients  have  special  values,  a  function  of  the  roots  may  be  formally 
altered  by  a  substitution  and  yet  experience  no  change  in  numerical  value. 
Take,  for  instance,  the  equation  with  special  coefficients,  Xs  =  1.  If  w  is  one 
of  its  complex  roots,  we  may  write  «o  =  w,  «i  =  w2,  (t^  =  w3.  The  function 
«02<*i  is  altered  in  form  by  the  substitution  ((X0n-2(ti)  >  Dut  not  in  value  ;  for, 
«02«j  =  «22«0  =  w.  That  functions  of  «0,  «lt  «2>  m^y  have  different  forms, 
but  the  same  numerical  value  is  seen  also  in  the  equalities 


SUBSTITUTION-GROUPS  125 

>  We  have  seen  that  the  alternating  group,  regarded  as  a  sub- 
group of  the  symmetric  group,  has  the  alternating  function 
which  belongs  to  it  (§  100).  Similarly  the  cyclic  group,  re- 
garded as  a  sub-group  of  the  symmetric  group,  has  the  cyclic 
function  which  belongs  to  it  (§  101).  The  cyclic  function  still 
belongs  to  the  cyclic  group  when  the  latter  is  considered  as  a 
sub-group  of  a  sub-group  of  the  symmetric  group. 

The  function  a^  +  xs  —  x2  —  x4  belongs  to  the  group  1,  (1  3), 
(2  4)  when  this  group  is  taken  as  a  sub-group  of  1,  (1  3)  (2  4), 
(1  2)  (3  4),  (1  4)  (2  3),  but  the  function  no  longer  belongs  to 
that  group  when  considered  as  a  sub-group  of  the  symmetric 
group ;  for  the  substitution  (1 3)  occurs  in  the  symmetric  group, 
but  not  in  the  given  sub-group,  and  yet  (1  3)  leaves  the  func- 
tion unchanged.  When  we  say  that  a  function  belongs  to  a 
group,  but  do  not  mention  of  what  other  group  the  given  group  is 
a  sub-group,  we  shall  understand  that  it  is  under  the  symmetric. 

112.    To  find  Functions  which  belong  to  a  Group.     Let  Ol  be 

a  sub-group  of  G,  G  being  of  the  degree  n,  and  let  a1;  a2>  •••>  <*-n 
be  distinct  quantities.  Let  also 

P  =/(«i,  —)  O 

be  a  rational  function  which  may  have  rational  coefficients  and 
which  will  assume  a  different  value  for  every  substitution  of 
the  group  G.  If  the  order  of  the  sub-group  GI  is  m,  we  obtain, 
on  operating  upon  p  with  the  substitutions  in  G^  m  distinct 

values>  fc*,j»",jv»  l 

If  now  we  operate  upon  the  functions  I  by  any  substitution 
in  Gl}  these  quantities  are  merely  permuted  among  themselves ; 
for,  any  value  p  thus  obtained  as  the  result  of  two  substitu- 
tions, Sj  and  s2,  of  the  sub-group  G»  is  the  same  as  that  obtained 
from  p  by  the  simple  substitution,  s3  =  s^  •  s2,  of  this  sub-group. 

These  facts  point  to  the  unexpected  conclusion  that,  in  the  theory  under 
development,  the  equation /(z)  =0  may  represent  a  more  general  case  when 
the  coefficients  are  particular  numbers  than  when  they  are  variables.  See  §  2. 


126  THEORY   OF    EQUATIONS 

If,  however,  we  apply  to  the  functions  I  a  substitution  of  G 
which  does  not  occur  in  Gly  we  obtain  a  series  of  functions 

P')  P'I)    '")  P'm-l) 

of  which  at  least  p'  does  not  occur  in  I.  For,  if  p'  did  occur 
in  I,  we  would  have  two  identical  functions,  distinct  from  p, 
resulting  from  the  application  to  p  of  two  different  substitu- 
tions. This  is  impossible. 

If  now  we  form  a  new  function  \j/  thus, 


where  t  is  a  variable,  it  is  evident  that  \f/  remains  invariant 
when  operated  on  by  the  substitutions  of  the  sub-group  GI,  but 
varies  for  any  substitution  in  G  which  does  not  occur  in  G^ 
Hence  if/  is  a  function  which  belongs  to  G1}  taken  as  a  sub-group 
of  G. 

We  are  at  liberty  to  assign  to  t  any  rational  value  which  will 
keep  \l/  distinct  from  any  value  obtained  for  it  by  application 
to  i/'  of  a  substitution  in  G  that  is  not  in  Gv.  One  such  value  is 

c«0. 

113.  This  method  of  finding  functions  belonging  to  a  group 
does  not  usually  furnish  simple  results  directly,  as  will  be  seen 
from  the  following  example. 

Ex.  1.    Form  a  function  of  «i,  02,  aa,  «4>  which  belongs  to 

(?2(4)  =  1,  (1  3)  (2  4),  taken  as  a  sub-group  of 
<?4WII  =  1,  (13)  (2  4),  (12)  (3  4),  (14)  (2  3). 

Assume  p  =  Ci«i  +  c2a.2  +  c3«3  +  c4«4,  such  that  p  assumes  four  distinct 
values  for  the  substitutions  of  Cr4(4>II.  The  substitutions  of  (r2<4)  applied 


p  =  Cl«l  +  C2«2  +  CS«3  +  C4«4, 
Pi  =  Ci«3  +  C2«4  +  C3«i  +  C4«2, 

hence    \f,  =  (t-p)(t-  p\)  =P-  («i  +  «3)  (tei  +  «cs)  -  (02  + 

+  («l2  +  «8a)CiC8  +  («22  +  «42)  C2C4 
+  «l«3(Cl2  +  C32)  +  «2«4(C22  +  C42) 
+  («2«3  +  <Xl«4)(ClCa  +  C3C4)  +  («l«2  +  «3«4)(ClC4 


SUBSTITUTION-GROUPS  127 

^  is  a  required  function.    By  inspection  we  see  that  ty  is  composed  of 
parts  which  are  themselves  functions  of  the  kind  sought  for.    These  parts  are 


tc3~)  -  C«2 

(«12  +  «32)ClC3  +  («22  +  «42)C2C4, 
«l«3(Cl2  +  C32)  +  a2«4(C22  +  C42). 

For  t  =  1,  ci  =  cs  =  —  1  and  c2  =  c4  =  +  1  we  obtain  the  simpler  form 

«i  +  «s  —  «2  —  «4- 
For  £  =  0,  Ci  =  CB  =  1,  c2  =  c4  =  z,  we  obtain  the  simpler  forms 

«12  +  «32  -  «22  -  «42, 


Ex.  2.   Assuming  p  =  «i  —  «2  +  i«3,  derive  functions  which  belong  to 

as  a  sub-group  of  6!6<3>. 
Taking  «  =  0,  we  get  (i  -  2)(«i«32  +  cc3a22  +  «2«12)  +  (i  +  2)(<x2«s2 
2  +  «i«22).      Then    show  that    «i«32  +  «3a22  +  «2«i2    and 
+  «i«22  each  belong  to  (?8<8>. 


*  Ex.  3.    Find  the  group  to  which  (cct  +  as)  («2  +  «4)  belongs. 
We  find,  by  trial,  which  of  the  substitutions  of  the  symmetric  group  of 
the  fourth  degree  leave  the  function  unaltered.     These  substitutions  are 


(«i«4«3«2).  These  substitutions  constitute  the  required  group.  From 
§  104  it  is  seen  to  be  Crs(4)-  From  the  behavior  of  this  group  toward  the 
given  function,  show  that  the  group  is  imprimitive. 

Ex.  4.  Find  the  group  to  which  «!«2  +  «3«4  —  («!«3  +  «2«4)  belongs. 

Ex.  5.    Find  the  group  to  which  (a  —  «i)(«2  —  «s)  belongs. 

Ex.  6.   Find  the  group  to  which  (ai«2  —  «3a4)2(«i«3  +  «2«4)2  belongs. 

Ex.  7.    Prove  that  the  substitutions  which  leave  unaltered  a  function 
of  n  distinct  letters,  form  together  a  group  of  the  wth  degree. 

*  Ex.  8.  Show  that  a^ad1  +  a-fatf  +  ----  h  «n-ip««»  +  ««"«i«,  where  p 
and  q  are  distinct  positive  integers,  is  a  cyclic  function. 

Ex.  9.    By  inspection  show  that  {(«  —  «2)  +  i  («i  —  fts)}2  belongs  to 
6!2(4)  as  a  sub-group  of  6r24(4).     Compare  with  Ex.  1. 

AC1     AT) 

Ex.  10.   Show  that  the  cross-ratio  of  four  points  (§  78)  k  =  =£-  +  -_—  , 

B(j      If  I) 

when  k  is  not  equal  to  —  1  or  to  w,  is  a  function  which  belongs  to 


128  THEORY    OF    EQUATIONS 

that  it  has  then  six  distinct  conjugate  values  ;  that  when  k  =  —  1  or  k  =  w, 
the  conjugate  values  are  formally  different  ;  that  the  numerical  values  coin- 
cide in  pairs  when  k  =  —  1,  and  in  triplets  when  k  =  w,  w  being  a  complex 
cube  root  of  —  1.  See  §  111. 

Ex.  11.  Find  the  values  of  the  roots  of  a;4  —  a;3  —  x  +  1  =  0,  and  show 
that,  for  these  values,  the  function  «'2«i  +  «i2«2  +  «22<*3  +  <*32<*  does  not 
belong  to  the  cyclic  group,  although  this  function  is  formally  altered  by 
all  substitutions  in  GW4*  which  do  not  occur  in 


*  Ex.  12.    Show  that,  for  the  general  quartic,  the  following  functions 
belong  to  the  cyclic  group  : 

(«  +  2  «i)  («i  +  2  «2)(«2  +  2  «3)(«3  +  2  a), 
a3«i(a2  +  2  «i)  +  aisa2(ai2  +  2  «2)  +  a^a.3(a-i2  +  2  «3)  +  «33a  (as2  +  2  a). 


CHAPTER   XII 

RESOLVENTS  OF  LAGRANGE 

114.  Resolvents.      Expressions,   known    as    "resolvents   of 
Lagrange,"  are  of  great  importance  in  researches  on  the  alge- 
braic solution  of  equations.     The  term  resolvent  is  used  in  two 
different  senses :   first,  to  represent  certain  auxiliary  equations 
used  in  the  resolution  of  given  equations ;  second,  to  represent 
certain  functions  used  in  the  resolution  of  equations.      The 
Lagrangian  resolvents  are  of  the  latter  kind ;  they  are  functions 
of  roots  of  unity  and  the  roots  of  the  given  equation. 

115.  Definition.      Let  f(x)  =  0  be  an  equation  having  the 
roots  a,  «!,  •••,  «„_!•     Let  w  be  any  one  of  the  nth  roots  of 
unity,  and  let  the  function  [<o,  a]  be  defined  as  follows : 

[w,  a]  =  a  -f-  ft>#i  +  to2«2  +  •••  +  w^Xi-i-  I 

The  expression  I  is  a  Lagrangian  resolvent. 

116.  Roots  expressed  in  Terms  of  Resolvents.     If  we  write 
the  Lagrangian  resolvents, 

[o>,  a]  =  a  +  <!)«!  +  <»2«2  +  •••  +  <">B~1«*-u 
[(•>!,  a]  =  a  +  (•>!«!  4-  tofa2  +  •••  +  wi""1^-!, 

O»-i,  a]  =  a  +  "«-i«i  +  w«-i2«2  H h  <«1,-i"~1«» 

and  add  them,  we  get      2  [to,  «]  =  na,  II 

where  1  signifies  the  sum  of  all  the  [o>,  a],  obtained  by  writing 
in  succession  w,  toj,  to2,  •••,  («)„_!  in  place  of  w. 
K  129 


130  THEORY   OF   EQUATIONS 

If  we  multiply  the  equations  in  I  by  o>~*,  wj"*,  •••,  <on__!~*, 
respectively,  and  then  add,  we  have  the  more  general  result, 

2w~*  [to,  «]  =  nak.  Ill 

Hence,  if  we  are  given  the  values  of  the  Lagrangian  resolvents 
of  an  equation  /(x)  =  0  of  the  «th  degree  and  the  nth  roots  of 
unity,  the  equation  /(x)  =  0  is  solved. 

117.  Theorem.  If  we  operate  upon  the  subscripts  of  a  in 
[to,  a]  with  the  cyclic  substitution  (0  1  2  3  •••  (n  —  1)),  [to,  a] 
becomes  w"1  [to,  a]  ;  if  we  operate  with  (0  1  2  •  •  •  (n  —  1))*,  [to,  «] 
becomes  a>~*  [to,  a]. 

If  we  operate  upon 

[<l>,    «]  =  «+<!>«!  H  ----   +  lO^Xi-l 

with  the  substitution  (0  1  2  •••  (n  —  1))  and  observe  that 
w""1  =  to"1,  etc.,  we  get 

2 


o>« 


Operating  in  this  manner  k  times,  we  can  easily  establish  the 
truth  of  the  second  part  of  the  theorem. 

118.   Theorem.     If  with  the  cyclic  substitution 
(012  ...  (n-1)) 

we  operate  upon  the  subscripts  of  a.  in  [to,  «],  the  subscript  of  the 
coefficient  of  each  power  of  to  in  [to,  a]"  undergoes  the  cyclic  sub- 
stitution (0  1  2  ...  (n  —  1))",  v  being  any  positive  integer. 

By  the  Polynomial  Formula  expand 

[to,  a]v  =  (a  +  w«H  -----  \-  aj"-1aB-i)l'» 

and  by  the  relation  o>n  =  1  reduce  all  exponents  of  to  to 
exponents  less  than  n.  Then  combine  all  terms  having  like 
powers  of  to.  We  get 


RESOLVENTS   OF   LAGRANGE  131 

t 

where  AQ,  Alf  •  ••,  An-l  are  expressions  of  the  degree  v  with  re- 
spect to  a,  «!,  «2,  -;  «n-i,  and  have  integral  numerical  coefficients. 
If  in  formula  I  we  replace  w  by  w,  Wl,  o>2,  •••,  wn_!  in  succes- 
sion, we  get  the  following  n  formulae : 

[CD,  «]"  =  A0  +  wAl  +  of  A^  +•••  +  co"-1^,.!, 

[«„    «]"  =  A  +  "1 A  +  toMs  +   ' ' '    +  <"' A-l, 
[<"«-l»   «]"  =  A  +  W»-lA  +  <->»-Mj  -i 1-  "W^A-l-  . 

It  was  shown  in  §  69,  Ex.  5,  that  the  sum  of  the  pfh  power  of 
the  nth  roots  of  unity  is  n  or  0,  according  as  p  is  divisible  or 
not  divisible  by  n.  Remembering  this  and  multiplying  the  n 
expressions  in  II  by  or*,  wj~*,  •••,  w,,_r*,  respectively  (k  being 
any  integer),  we  get,  after  adding  the  n  resulting  expressions, 

ta 

nAk  =  2a>~*  •  [<o,  a]",  III 

where  ^  indicates  the  sum  of  all  the  expressions  obtained  by 
writing  in  succession  w,  wj,  w2,  •••,  (an_l  in  place  of  w.  If  now  we 
operate  upon  the  subscripts  of  a,  occurring  in  each  of  the  v  fac- 
tors [<o,  a]  in  the  right  member  of  III  with  the  cyclic  substitu- 
tion (0  1  2  ...  n  —  1),  we  get,  §  117, 

2<o-*-"  •  [o>,  a]".  IV 

Now,  by  writing  k  +  v  for  k  in  formula  III,  we  obtain 

2w~*""[a),  a]"  =  nAk+v. 

In  other  words,  the  substitution  (01  2  •••  (n  —  1)),  applied 
to  the  subscripts  of  a  in  the  right  member  of  III  causes  Ak 
to  be  replaced  by  Ak+v.  But  Ak  is  transformed  directly  into 
Ak+v  by  the  application  to  its  subscript  of  the  substitution 
(0  1  2  •••  (n  —  1))".  Hence  the  theorem  is  established. 

Ex.  1.  Illustrate  this  theorem  by  the  roots  «0»  <*i>  «2  of  the  cubic, 
taking  v  —  2. 

We  have  [w,  «0]  =  «o  +  ««i  +  <»2«2, 

[w,  #o]2  =  -4o  +  -4iw  +  A^u"2, 
where    ^10  =  «o2  +  2  a^,  .4i  =  «22  +  2  Oottj,  .42  =  «i2  + 


132  THEORY   OF   EQUATIONS 

Operating  upon  the  subscripts  of  «  in  [w,  «]  by  (0  1  2),  we  get 


and  («i  +  w«2  +  w2«o)2  =  A2  +  A0w 

We  see  that  AO,  A\,  A%,  when  operated  ou  by  (0  1  2)2,  become  respec- 
tively Az,  AO,  A\. 

Ex.  2.  Illustrate  this  theorem  by  taking  v  =  3  in  Ex.  1,  and  show  that 
the  function  belongs  to  the  cyclic  group. 

Ex.  3.  Show  that  (0  1  2),  applied  to  the  subscripts  of  «0,  «i,  (t-z,  in 
[w2,  «]2  =  («o  +  w2«i  +  w4«2)2  =  AO  +  A\w  +  Azu2,  produces  the  same 
effect  as  (0  1  2)4  applied  to  the  subscripts  of  AO,  A\,  A*. 

Ex.  4.  Show  that  (012  3)  applied  to  the  subscripts  of  «0,  «i,  «2,  (t-z, 
in  [w3,  «]2=  («o  +  w3«i  +  ^6«2  +  w9«s)2  =  A  +  AIW  +  ^lew2  +  ^w3,  where 
w  =  —  i,  produces  the  same  effect  as  (0  1  2  3)  6  applied  to  the  subscripts 
of  AQ,  AI,  AZ,  AS. 

119.   Theorem.     If  with  the  cyclic  substitution 
(012-..  0-1)) 

we  operate  upon  the  subscripts  of  a,  the  subscript  of  the  coefficient 
of  each  power  of  w  in  the  product  of  [o>,  «]"  •  [o/i,  «]"i  •  [G/J,  «]"» 
•--suffers  the  substitution  (0  1  2  •••  (n  —  l))v+Vi+Vi+—  ,  where 
v,  VD  i/2>  ••*  are  positive  integers  and  Xi}  \2,  •••  positive  or  negative 
integers. 

This  theorem  is  a  generalization  of  the  preceding  and  is 
proved  in  the  same  way.  The  product  yields  the  equality 

[w,  a]"  •  [a/i,  a]  •'i  •  [wA2,  a]l'i  •••  =  £0  +  ^A  +  w2^  + 
...  +a)»-1B,,-iJ 

where  50,  5j,  •••,  #„_!  are  functions  of  the  roots  a,  «j,  •»,  a,,_a. 
Replacing  w  successively  by  w,  o^,  o)2,  -.-,  o>n_!,  we  have  all 
together  n  expressions.  Multiply  them  by  w  *,  o^"*,  w2~*,  ••• 
respectively,  then  add  the  resulting  products,  and  we  get 

nBk=  iioT'O,  a]"  •  [<A  «]"i  ....  I 


RESOLVENTS    OF    LAGEANGE  133 

» 

To  the  subscripts  of  a  in  the  right  member  of  I  apply  the 
substitution  (0  1  2  •••  (n  —  1)),  and  we  get 

SoT*-"-*!*!-  -[a,,    «]"    •    [o/l,    «]*!    ..., 

which  expression  is  recognized  by  I  to  be  equal  to  nBt+v+  ^  + .... 
But  Bk  is  replaced  by  Bk+v+vi^l+ ...,  if  we  operate  upon  Bk  with 
the  substitution  (0 1  2  •••  (n  —  l))"+^i+-.  Hence  the  theorem 
is  established. 

*  Ex.  1.    Show  that  the  function  [w,  «]n  belongs  to  the  cyclic  group  of 
the  degree  n. 

If  we  operate  upon  [w,  «]  with  any  such  substitution  (0  1  2  •••(»  —  1)) 
of  the  cyclic  group,  the  effect  is  the  same  upon  the  coefficients  Bk  of 
[w,  cc]n  as  if  the  substitution  (012  •••  (n  -  1))"  were  applied  to  the 
subscripts  of  Bk  directly,  §  118.  But  (012  •••  (M  -  1))"  is  the  iden- 
tical substitution ;  hence  it  brings  about  no  change.  Consequently 
[w,  «]"  is  invariant  for  the  cyclic  group.  This  invariance  holds  for  no 
substitution  of  the  symmetric  group  of  degree  n,  except  the  substitutions 
which  occur  also  in  the  cyclic  group.  Hence  [w,  a]n  belongs  to  the  cyclic 
group. 

*  Ex.  2.    Show  that  the  product  [w,  a]"-*  .  [w\  «]  belongs  to  the 
cyclic  group  of  degree  n. 

By  §  118,  IV,  the  cyclic  substitution  (0  1  2  •••  n  —  1),  effected  upon 
the  subscripts  of  a  in  [a»,  «]"-*  gives  w-n+*  [w,  a]"-*.  When  operated 
upon  those  in  [u>A,  «]  it  gives  W~A[WA,  «].  Hence,  when  operated  upon 
the  product  of  the  two,  we  get  a»-"+^-^[w,  a]n~*  •  [w\  a],  where 

(j~n+A — A  —  tt)~n  —  1. 

Ex.  3.  Show  that  (a  —  ia\  —  «2  +  i«s)4  belongs  to  the  cyclic  group  of 
degree  four. 

For  convenience,  let  —  i  =  w,  and  we  have  («  +  w«!  +  w2«2  +  «3«s)4, 
which,  by  §  118,  IV,  becomes  w~4(a  +  w«i  +  w'2«2  +  w3«s)4  when  operated 
upon  by  (0  1  2  3). 

Ex.  4.     Notice  if  the  following  functions  belong  to  the  cyclic  group  of 

degree  four : 

(«  +  t«!  -  «2  -  zas)«, 

(a  —  i'«!  —  «2  +  i«3)  (a  +  iaj  —  «2  —  i'«8), 
(a  —  «!  +  «2  —  «3)  («  —  i«i  —  «2  +  i««)2, 
(a  -  «!  +  «2  -  as)2- 


CHAPTER   XIII* 

THE  GALOIS  THEORY  OF  ALGEBRAIC  NUMBERS.    REDUCTIBILITY 

120.  Definition  of  Domain.  A  set  of  numbers  is  called  a 
domain  of  rationality  or  simply  a  domain,  when  the  sums,  dif- 
ferences, products,  and  quotients  of  any  numbers  in  the  set 
(excluding  only  the  quotients  obtained  through  division  by  0) 
always  yield  as  results  numbers  belonging  to  the  set. 

All  rational  numbers  (integers  and  rational  fractions,  taken 
both  positively  and  negatively)  constitute  such  a  domain,  for 
this  system  of  magnitudes  is  complete  in  itself  in  the  sense 
that  any  of  the  four  operations  involving  any  of  these  numbers 
never  yields  as  a  result  a  number  which  does  not  belong  to 
the  set. 

The  integers  by  themselves  do  not  constitute  a  domain,  for 
the  quotient  of  two  integers  may  be  fractional. 

All  the  numbers  of  one  domain  O  may  be  contained  in  a 
second  and  larger  domain  O'.  In  this  event  the  smaller  domain 

0  is  called  a  divisor  of  the  other  O',  and  O'  is  called  a  domain 
over  ft. 

For  example,  the  complex  numbers  of  the  form  a  +  ib,  where 

1  =  V—  1  and  a  and  b  signify  rational  numbers,  are  a  domain 
of  which  the  domain  of  rational  numbers  is  a  divisor. 

Another  example  of  domains  of  numbers  is  the  one  embrac- 
ing all  real  numbers,  whether  rational  or  irrational.  Still 
another  is  the  domain  consisting  of  all  numbers,  a  +  ib,  where 
a  and  6  are  fational  or  irrational. 

*  In  the  exposition  of  the  Galois  theory  in  this  and  the  succeeding  chapters 
we  have  followed  the  treatment  given  by  H.  Weber  in  his  Lehrbuch  der 
Algebra,  Vol.  I,  pp.  491-698. 

134 


GALOIS   THEORY   OF   ALGEBRAIC    NUMBERS          135 

121.  The  Domain  O(1).      The  domain  of  rational  numbers  is 
a  divisor  of  all  domains,  for  each  domain  contains  at  least  one 
number  n  different  from  0 ;  hence  it  contains  also  n  -=-  n  or  1. 
But  if  unity  belongs  to  the  domain,  then  it  -embraces  all  num- 
bers obtained  by  addition  and  subtraction  of  units,  that  is,  all 
positive  and  negative  integers ;  from  the  latter  we  can  by  divi- 
sion derive  all  rational  fractions.     Hence  the  rational  numbers 
occur  in  every  domain.     Hereafter  we  shall  indicate  the  domain 
of  rational  numbers  by  O(1). 

122.  Adjunction.      Let  fl  signify  any  domain.      If  we  add  to 
it  any  number  a  which  does  not  already  belong  to  it,  then  the 
new  system  of  numbers  does  not  constitute  a  domain  unless  we 
add  also  all  numbers  arising  from  a  finite  number  of  additions, 
subtractions,  multiplications,  and  divisions  involving  a  and  all 
numbers  in  the  domain  O.     Let  us  designate  the  new  domain 
thus  obtained  by  O(a).     It  is  evident  that  O  is  a  divisor  of  O(a). 

This  process  of  obtaining  the  domain  O(a)  from  O  is  called 
adjunction.  We  say  that  we  adjoin  a  to  li  and  obtain  Ii(a).  By 
the  adjunction  of  i  to  the  domain  of  rational  numbers  O(1)  we 
obtain  the  domain  of  complex  numbers  O(li<).  This  embraces 
all  numbers  of  the  kind  a  +  ib,  where  a  and  b  have  rational 
values.  In  general,  if  we  adjoin  a  to  fl(1),  we  get  fi(li  a). 

Ex.  1.  Show  that  the  rational  (proper)  fractions  do  not  constitute  a 
domain. 

Ex.  2.    Show  that  0  satisfies  the  definition  of  a  domain. 

123.  Reducibility  Defined.      Let  the  integral  function 

/(a?)  =  a0o;n  +  a^'1  -\ \-  a^x  +  an 

have  coefficients  a0,  a1?  ••-,  a«,  all  of  which  belong  to  some 
domain  O.  Then  we  shall  say  that  f(x)  is  a  function  in  O  and 
/(»)  =  0  is  an  equation  in  O.  If  the  function  f(x),  in  which  n 
is  some  integer  >1,  can  be  decomposed  into  factors  of  lower 


136  THEORY    OF    EQUATIONS 

degree  with  respect  to  x,  such  that  the  coefficients  of  the  fac- 
tors are  numbers  belonging  to  the  domain  O,  then  the  function 
f(x)  is  called  reducible  in  O  ;  otherwise  it  is  called  irreducible 
in  O. 

Thus,  if  J3  designates  the  domain  of  rational  numbers, 
then  x2  —  y2  is  reducible  in  O,  because  it  yields  the  factors 
(x  +  y)  (x  —  y).  On  the  other  hand,  x2  —  3  y'2  is  irreducible  in 

O,  because  some  of  the  coefficients  of  its  factors 

»  •» 


are  not  rational. 

If,  however,  we  form  a  new  domain  by  the  adjunction  of 
«  =  V3  to  the  domain  of  rational  numbers,  we  obtain  (i(1)  a), 
embracing  numbers  of  the  kind  a  +  V3  b,  where  a  and  b  are 
rational.  With  respect  to  this  larger  domain  the  functions 
x2  —  y2  and  x2  —  3  y2  are  on  an  equal  footing,  for  both  are 
reducible  in  O(li  a),  since  the  coefficients  of  the  two  factors  of 
each  function  are  numbers  belonging  to  the  same  domain  O(1)  a). 

Ex.  1.  Find  out  which  of  the  following  functions  are  reducible  in  the 
domain  of  rational  numbers  tyi)  : 

(a)  y?  +  2  x  +  1, 
(6)   x4  +     x2  +  l, 

(c)  a;*  +     x  _  i, 

(d)  x2+     z  +1, 

(e)  cc2  +  l. 

Ex.  2.  For  each  of  the  above  functions  which  are  irreducible  in  fl(i), 
find  by  adjunction  the  smallest  new  domain  in  which  the  function  is 
reducible. 

Ex.  3.  Find  a  domain  such  that  all  the  functions  of  Ex.  1  will  be 
reducible  in  it. 

124.  Algebraic  Numbers.  All  numbers  which  are  roots  of  an 
algebraic  equation 

#   =  OOJ"      axn~l       •••       a-x      a   =  0 


GALOIS    THEORY    OF    ALGEBRAIC    NUMBERS          Io7 

with  integral  coefficients  are  called  algebraic  numbers.  Numbers 
which  cannot  occur  as  roots  of  an  algebraic  equation  are  called 
transcendental.  It  was  first  proved  by  Hermite  (1873)  that  e, 
the  base  of  the  natural  system  of  logarithms,  is  a  transcen- 
dental number.  In  1882  Lindemann  first  demonstrated  that 
TT,  the  ratio  of  the  circumference  of  a  circle  to  its  diameter,  is 
also  transcendental.  If  to  the  domain  of  rational  numbers 
O(1)  we  adjoin  TT,  we  obtain  a  transcendental  domain.  If  the 
number  adjoined  to  O(J)  is  algebraic,  the  new  domain  is  called 
an  algebraic  domain. 

125.  Irreducible  Equations.     An  equation, /(cc)  =  0  is  said  to 
be  reducible  or  irreducible  in  a   domain  O,  according  as  the 
function /(#)  is  reducible  or  irreducible  in  O. 

If  we  adjoin  to  the  domain  O  one  of  the  roots  a  of  the 
equation  f(x)  —  0,  then  if  a  does  not  belong  to  the  domain  fi, 
we  obtain  a  new  domain  O(a)  which  is  an  algebraic  domain 
over  O. 

126.  Theorem.     If  f(x)  =  0  and  F(x)  =  0  are  both  equations 
in  the  domain  Q,  and  if  f(x)=  0  is  irreducible  in  O  and  has  one 
root  which  satisfies  F(x)  =  0,  then  all  its  roots  satisfy  F(x)  =  0. 

Since  the  two  equations  have  at  least  one  root  in  common, 
the  two  f unctions /(x)  and  F(x)  have  a  common  factor  involv- 
ing x.  But  we  know  that  the  highest  common  factor  is  found 
by  ordinary  division,  i.e.  by  a  process  which  nowhere  intro- 
duces numbers  not  found  in  the  given  domain  of  rationality. 
The  highest  common  factor  is  therefore  a  function  in  O. 
But  /(«),  being  irreducible,  has  no  factor  in  O  involving  x, 
except  itself.  Hence  the  highest  common  factor  must  be  either 
f(x)  or  a  quantity  differing  from  f(x)  by  a  constant  number. 
In  other  words,  we  must  have  either  F(x)  =  c  •  f(x)  or 
F(x)  =  g(x)  •  f(x),  where  g(x)  is  a  function  in  ft. 

Ex.  1.  The  cubic  x8  —  2  x2  —  x  +  1  —  0  has  three  incommensurable 
roots  and  is  therefore  irreducible  in  the  domain  n(1j.  It  has  one  root  in 


138  THEORY    OF   EQUATIONS 

common  with  x*  -  3  x3  +  x2  +  2  x  -  1  =  0.  Find  the  H.  C.  F.  of  the  two 
functions  and  show  that  all  the  roots  of  the  first  equation  satisfy  the 
second. 

Ex.  2.  The  function  x2  +  6x  +  7  is  irreducible  in  fi(i),  and  it  is  not  a 
divisor  of  x8  +  3  x2  +  3  x  4-  1.  From  these  data  show  that  the  two 
fuuctions  cannot  have  a  common  factor. 

Ex.  3.  The  equation  ax2  +  bx  +  c  =  0  in  ft(i)  has  a  root  in  common 
with  x3  +  5  x2  +  10  x  +  1  =  0.  Show  that  a  -  b  =  c  =  0. 

Ex.  4.  Prove  that  two  functions  in  ft,  0(x)  and  /(x),  cannot  have  a 
common  factor  which  is  a  function  of  x  in  0,  if  /  (x)  is  irreducible  and  not 
a  divisor  of  $(x). 

Ex.  5.  If  a  root  of  the  irreducible  equation  /  (x)  =  0  in  ft  satisfies  the 
equation  0(x)  =  0  in  fi,  and  if  /(x)  is  of  higher  degree  than  0(x),  then  all 
the  coefficients  of  <£(x)  must  be  zero. 

127.  Gauss's  Lemma.  If  f(x)  has  integral  coefficients  and  can 
be  resolved  into  rational  factors,  it  can  be  resolved  into  rational 
factors  with  integral  coefficients. 

Consider  the  two  functions,  , 


Let  k  be  the  H.  C.  F.  of  the  integers  a0,  a1}  a2,  •••  ;  and  let  I  be 
the  H.  C.  F.  of  the  integers  b0,  bl}  b2,  •••. 

Also  let  k  be  relatively  prime  to  ra,  and  let  I  be  relatively 
prime  to  n. 

We  may  now  write 

G(x)  =  k  .  g(x),          H(x)  =  I  •  h(x), 

where  g(x)  and  h(x)  are  functions  whose  denominators  are, 
respectively,  m  and  n.  The  numerator  of  g(x)  is  an  integral 
function  of  x  with  integral  coefficients  which  have  no  common 
factor,  except  1.  The  same  is  true  of  the  numerator  of 
h(x).  Hence  the  smallest  denominator  of  the  product  g(x)  •  h(x) 
is  raw. 


139 

Consider  the  case  when  the  product  G(x)  •  H(x)  has  only 
integral  coefficients.  Then  it  is  evident  that  k  •  I  must  be 
divisible  by  m  -  n.  Since  Jc  is  relatively  prime  to  m,  and  I  to  n, 

it  follows  that 

K  =  pn,         I  =  qm, 

where  p  and  q  are  integers.     We  may  now  write 


where  the  functions  g\(x)  and  h^x)  have  only  integral  coefficients. 
Consequently,  if  f(x)  is  resolvable  into  two  rational  factors 
G(x)  and  H(x),  which  have  fractional  coefficients,  so  that 


then  we  have  also      f(x)  =  pq  •  g^x)  •  \(x), 

where  the  coefficients  are  integral  throughout.  Hence,  if  f(x) 
is  resolvable  into  rational  factors,  it  is  resolvable  into  such 
factors  with  integral  coefficients. 

128.  Reducibility  of  /(a?).  Whether  the  function  /(«),  in 
which  the  coefficients  are  integers  and  the  degree  n  does  not 
exceed  4  or  5,  is  reducible  or  not  in  the  domain  O(1),  can  readily 
be  ascertained  by  the  aid  of  Gauss's  lemma  and  ordinary 
algebra. 

We  assume  that,  in  /(a/),  the  coefficient  a0  of  xn  is  unity.  If 
«o  is  not  unity,  we  can  change  the  function  so  that  it  will  be 

unity  by  taking  x  =  ^-,  and  multiplying  by  ann~l. 
a0 

For  every  integral  value  a  of  x,  which  causes  /(a?)  to  vanish, 
we  have  a  factor  x—  a,  of  /(#),  §  3.  Here  «  must  be  a  factor 
of  an.  This  consideration  enables  us  always  to  determine  the 
reducibility  or  irreducibility  of  functions  f(x)  of  the  second  or 
third  degree. 

If  f(x)  is  of  the  fourth  degree,  then,  if  there  is  no  linear 
rational  factor,  there  can  be  no  cubic  rational  factor.  To  test 


140  THEORY   OF    EQUATIONS 

for  quadratic  rational  factors,  divide  x4  +  a^  +  afl?  +  aax  -f-  a4 
by  x2  +  ax  -f-  ft,  where  a  and  (3  are  integers  to  be  determined, 
if  possible.  That  there  may  be  no  remainder,  we  must  have 

«a  —  <hft  +  «/?  =  «(«2  —  ft  —  2 


Hence  a  =  «*f*-"fm  U 

a^-  ft2 

We  have  the  rule  :  See  whether  any  factor  ft  of  a4  makes  a  an 
integer  in  II.  If  a  and  ft  are  such  integers,  which  also  satisfy  I, 
then  x2  -\-  ax  +  ft  is  a  rational  factor  sought. 

Similarly,  if  f(x)  is  of  the  fifth  degree.  First  search  for 
linear  rational  factors  x  —  c.  If  none  are  present,  there  is  no 
quartic  rational  factor.  Look  for  a  quadratic  rational  factor 
x2  +  ax  +  ft.  If  quadratic  factors  are  likewise  absent,  there 
can  be  no  cubic  rational  factor,  and  the  function  is  irreducible. 

Dividing  Xs  -j-  a^x*  +  a.p?  +  a^x?  +  a4x  +  «5  by  x2  +  ax  +  ft,  we 
get  as  the  conditions  for  zero  remainder, 


=  a(a3  —  aift  +  2aft  —  a2a  +  c^er  —  a3), 
%  =  ft(as  -alft  +  2aft-  a2a  +  aX  -  a8).  Ill 

Whence  a  =  ~  c*  ±  Vc^  ~ 

2c0 

where  c0  =  /82, 


c2  =  Oj/82  -  a4^8  -  yS3. 

If  )8  is  a  factor  of  a5,  if  «  is  an  integer,  and  III  is  satisfied, 
then  x2  +  ax  +  ft  is  a  factor  sought. 


Ex.  1.    Is/(se)  =x5  +  4x*  +  4x3  +  9a;2  +  8x  +  2  reducible  in 

Since  /(x)  does  not  vanish  for  x  =  ±  1  or  ±2,  there  are  no  linear  nor 

quartic  factors  in  Ity).     Take  /3  =  2,  then  Co  =  4,  Ci  =  —  14,  02=  —  8,  a=4. 

Condition  III  is  satisfied.     Hence  a;2  +  4  x  +  2  is  a  factor. 


o  .>  '0 


GALOIS    THEORY    OF    ALGEBRAIC    NUMBERS          141 


Ex.  2.    Are  the  following  reducible  i 

(1)  xs  +  2  x2  +  3  x  -  6.  (5)  x*  +  10  x3  -  100  x2  -  a;  +  1. 

(2)  x3  +  3  x2  +  8  x  -  2.  (6)  x5  +  x3  +  x2  +  x  +  7. 

(3)  x4  +  x3  +  x2  +  x  -  4.  (7)  x5  +  2  x4  +  3  x8  +  4  x2  +  3  x  +  2. 

(4)  x*  +  9  x3  +  25  x2  +  22  x  +  6.        (8)  x5  +  x  +  1. 

129.  Eisenstein's  Theorem.  Tf  p  is  a  prime  number,  and 
a0,  a1?  •••,  an  integers,  all  (except  a0)  divisible  by  p,  but  an  not 
divisible  by  p"2,  then  is  f(x)  =  a^xn  +  a-^""1  +  •••  +«„  irreducible. 

For,  if  /(#)  could  be  resolved  into  factors,  the  coefficients  of 
the  factors  could  be  integers.  We  could  have 

f(x)  = 
where 

Since  an  is  divisible  by  p,  but  not  by  p2,  and  an  =  CA  •  dt,  it 
follows  that  one  of  the  factors  CM  dk,  is  divisible  by  p,  but  not 
the  other.  Let  CA  be  the  factor  divisible  by  p.  Then  not  all 
the  coefficients  c  are  divisible  by  p,  else  a0  would  be  divisible 
by  p.  Let  c,  be  a  coefficient  not  divisible  by  js,  while  ce+1,  cc+2, 
•  ••  CA,  are  each  divisible  by  p.  The  coefficient  of  .T*~",  in  the 
product  of  the  two  factors  of  /(#),  is  then 


Since  every  term  in  this  polynomial  is  divisible  by  p,  except 
the  first  term,  the  polynomial  is  not  divisible  by  p.  But,  by 
assumption,  the  only  coefficient  of  f(x)  which  is  not  divisible 
by  p  is  OQ.  Hence  &~v  =  xn,  which  is  impossible,  since  h  must 
be  less  than  n. 

Ex.  1.    Show  by  §  129  the  irreducibility  of 

2  xs  +  9  x2  4-  6  x  -f  12, 
4  x5  +  14  x*  +  21  x  +  35. 


142  THEORY   OF   EQUATIONS 

yf  ^ 

130.    Irreducibility   of  —  •      When  p  is  a  prime  number, 

P      -\  x  —  1 

the  equation  x  ~     =  0  is  irreducible. 


If  in —  =  0,  we  put  x  =  z  4- 1,  then  expand  the  binomials 

and  simplify,  we  get 


Since  this  equation  is  irreducible  by  §  129,  the  given  equa- 
tion is  irreducible. 


131.  Exclusion  of  Multiple  Roots.      Unless  the  contrary  is 
specifically  asserted  we  shall  assume  in  what  follows  that  the 
equation  f(x)  =  0  has  no  multiple  roots.     This  can  be  done 
without  loss  of  generality.     For,  if  f(x~)  =  0  has  multiple  roots, 
we  can  divide  f(x)  by  the  H.  C.  F.  of  f(x)  and  f(x),  as  in  §  21, 
and  obtain  a  quotient  g(x).     Then  g(x)  =  0  is  an  equation  in  O, 
having  all  its  roots  distinct,  and  the  theorems  which  will  be 
given  apply  to  g(x)  =  0. 

Ex.  1.    Show  that  /(x)  =  0  is  reducible  if  it  has  multiple  roots. 

132.  Definition  of  Degree  of  a  Domain  and  of  Normal  Domain. 
If  the  irreducible  equation  /(«)  =  0,  having  «  for  one  of  its 
roots,  is  of  the  nth  degree,  the  domain  Q(a)  is  said  to  be  of  the 
nth  degree. 

Since  f(x)  =  0  is  irreducible  in  fi,  it  follows  that  none  of  its 
roots  belong  to  the  domain  12.  For,  if  the  root  a  were  a  num- 
ber in  the  domain  O,  x  —  a  would  be  a  factor  in  O,  and  /(a?) 
would  be  reducible.  It  is  evident  that  each  root  of  f(x)  =  0, 
when  adjoined  to  O,  gives  rise  to  a  domain  over  O,  §  120. 
Thus,  if  «,  «u  «2,  •  •  •,  «„_!  are  the  roots  of  f(x)  =  0,  we  obtain 
the  n  domains, 


GALOIS   THEORY    OF    ALGEBRAIC    NUMBERS          143 

The  domains  I  are  said  to  be  conjugate  to  the  O(a).  These 
domains  may  be  all  different  from  each  other ;  some,  or  all  of 
them,  may  be  alike. 

A  domain  which  is  identical  with  all  its  conjugate  domains  is 
called  a  normal  domain.  The  laws  of  normal  domains  are  far 
simpler  than  those  of  others.  The  great  advances  in  algebra 
made  by  Galois  rest  mainly  on  the  reduction  of  any  given 
domain  to  a  normal  domain. 

133.  Theorem.  Any  number  in  a  domain  O(a)  can  be  ex- 
pressed as  a  function  of  a  in  fl. 

By  definition  of  a  domain  (§  120)  any  two  numbers  in  it,  com- 
bined by  addition,  subtraction,  multiplication,  or  division,  yield 
a  number  occurring  in  the  domain ;  also  any  number  added  to 
or  subtracted  from  itself,  multiplied  or  divided  by  itself,  yields 
a  number  belonging  to-  the  domain. 

The  domain  O(a)  was  obtained  by  adjunction  of  a  to  n. 
Hence  the  numbers  in  O(a),  whether  occurring  in  O  or  not,  were 
obtained  by  carrying  out  the  four  operations  of  addition,  sub- 
traction, multiplication,  and  division  upon  a  and  the  numbers 
in  ft.  This  means  that  every  number  in  O(o)  is  expressible  as 
a  function  of  a  in  Q. 

Ex.  1.  Show  that  the  roots  of  x*  -  10  x2  +  1  =  0  define  a  normal 
domain. 

The  roots  are  a  =  \/2  +  V3,  «i  =  -  \/2  +  V3,  «2  =  -  ^2  -  V3» 
«3  =  V2  -  \/3.  We  have  «=-«2  =—  =  -  —  .  Hence  it  follows  that 

r>  «1  ft3 

PO,  a)  =  "(1,  aj)  =  "(1,  a,)  =  "(1,  <»„)• 

Ex.  2.  Show  that  the  domain  defined  by  the  roots  of  the  irreducible 
equation  a;3  +  x  +  1  =  0  is  not  normal. 

By  Descartes'  Rule  we  see  that  the  equation  has  only  one  real  root.  No 
complex  root  can  be  a  rational  function  of  a  real  root.  Hence  the  three 
domains  %,  a),  Q(i,  a  >,  fi(i,  „  )  cannot  be  identical  and  therefore  not  normal. 
But  the  two  domains  defined  by  the  complex  roots  are  the  same ;  for, 

8P  4-  "y2      TI 
if  p  +  iy  and   p  -  iy  are  the  complex  roots,  ft  -  iy  =  H       .'  •    Hence 

(3  -  iy  is  a  number  in  the  domain  obtained  by  adjoining  /3  +iy. 


144  THEORY    OF   EQUATIONS 

Ex.  3.   Show  that  the  roots  of  x4  —  22  x2  +  1  =  0  yield  a  normal  domain. 

Ex.  4.  Show  that  the  roots  of  an  irreducible  quadratic  determine  a 
normal  domain. 

Ex.  5.  Show  that  any  three  roots  of  x4  +  x8  +  x'2  +  x  +  I  =  0  are 
powers  of  the  fourth  and  that  the  domain  12(1,  a)  is  normal.  See  Ex.  2,  §  67. 

Ex.  6.  Express  as  a  function  of  V5  in  Q(i,  ,-)  the  following  numbers  of 
the  domain  fyi,^):  1,  10  i,  3  +  4V  —  5. 

Ex.  7.  Define  the  domain  fl  which  includes  the  number 
(V2  +  iVS  -  V6)-3. 

134.  Conjugate  Numbers,  Primitive  Numbers.  Suppose  a 
number  N  —  <£(«),  where  <f>  indicates  a  function  in  fi.  If 
a,  «j,  •••,  «»  are  the  roots  of  an  irreducible  equation  f(x)  =  0, 

then         N=4>(a),     N,  =  <£(«,),  ...,     Nn_l  =  tfa^,  I 

represent  n  numbers,  one  from  each  of  the  conjugate  domains, 


The  numbers  I  are  said  to  be  numbers  conjugate  to  N. 

Some  or  all  of  these  numbers  conjugate  to  .2V  may  be  equal  to 
each  other. 

A  number  JVin  the  domain  O(a),  which  is  different  from  all 
its  conjugate  numbers,  is  called  a  primitive  number  of  the 
domain.  Otherwise  it  is  called  imprimitive. 

135.  Primitive  Domains.  A  domain  I7(a)  is  called  primitive 
when  it  contains  no  imprimitive  numbers  except  the  numbers 
in  the  domain  O  ;  it  is  called  imprimitive  when  it  contains 
other  imprimitive  numbers  besides. 

Ex.    1.   The  equation  f(x)  —  x2  +  1  =  0  has  the  roots  ±  ?'.     Here  a  =  t 

a2  -f  «  +  2 
and  «i  =  —  i.    Let  us  assume  <t>  (a)  =  --  ,  then  <f>  («)  =  —  i  +  1  =  N, 

and  Ni  =  i+l.    Hence  JV,  being  unlike  JVi,  is  a  primitive  number  in  ft(i,  o- 
Next,  let  us  assume  </>(«)  =  «—  a  =  a\  —  «x  =  0.     Hence  0  is  an  imprimi- 
tive number  in  %,  ,> 


GALOIS   THEORY   OF    ALGEBRAIC    NUMBERS          145 

More  generally,  if  <f>  (i)  =  a  +  ib,  where  a  and  6  are  rational  numbers, 
then  0(  —  i)  =  a  —  ib  ;  if  0(i)  =^-  =  a,  then  0(—  i)  =  <k,  Hence  the  im- 

primitive  numbers  are  in  this  example  confined  to  those  that  are  rational, 
and  the  domain  Q^t  *)  is  primitive.  Since  both  fi(lj  ^  and  (i(it  _,-)  are 
domains  containing  numbers  a  +  z'6,  where  a  and  b  are  rational,  and  may 
be  positive  or  negative,  it  follows  that  the  two  conjugate  domains  are 
identical.  Hence  Q(it  ,•)  is  a  normal  domain. 

Ex.  2.    The  roots  of  the  irreducible  equation  x2—  2=0  are  ±  V2.     Show 

that  -  —  —  is  a  primitive  number  of  fl(1  v/2)>  that  10  is  imprimitive,  that 

V2 
the  domain  fl(li  v/§)  is  primitive  and  normal. 

Ex.  3.  If  a  is  a  root  of  xz  +  10  x  +  1  =  0,  define  the  functions  of  a  such 
that  JV  will  be  the  imprimitive  number  5. 

Ex.  4.  Show  that  the  number  jV  =  a2  +  a3,  belonging  to  the  normal 
domain  tyi,  a),  in  Ex.  2,  §  67,  is  imprimitive  and  that  the  domain  12(i,  a) 
is  imprimitive. 

Ex.  5.  If  2V=«2,  where  a  is  a  root  of  ^  +  1=0,  show  that  JVis  imprimi- 
tive, that  NI  =  a?  —  a  is  primitive,  that  the  domain  12(1,  a)  is  normal  and 
imprimitive. 

Ex.  6.  If  .ZV  =  a16  and  a  is  a  root  of  xs  +  1  =  0,  prove  that  N  is  imprimi- 
tive, that  fl(i,  a)  is  normal  and  imprimitive. 

Ex.  7.    If  a  is  a  root  of  x~  —  1  =  0,  prove  that  0(i,  a)  is  imprimitive.    / 

136.  Theorem.  Every  number  N  in  the  domain  nu)  of  the  nth 
degree  is  the  root  of  some  equation  of  the  nth  degree  in  to,  the 
other  roots  of  which  are  the  remaining  numbers  conjugate  to  N, 
viz.  JV,,  N2,  •",  JVn_!. 

Take  the  product 

(y-N^y-N,)  -.  (y-Nn  ,}  =  yn  +p<y»-l+  ^  +pn  = 
in  which       —  pl  =  N  +  N-^  -\  -----  \-  -ZVn_i, 


We  see  that  all  the  coefficients  pl9  p2,  •-,  pn  are  rational  sym- 
metric functions  of  the  numbers  N,Nlt  •••,  ^-i-  Since  N=4>(a), 
JV,  =  <£(«,)>  "'}  Nn-i  =  <K«»  0  (§  184)'  where  <f>  is  a  function  in 


146  THEORY   OF   EQUATIONS 

12,  it  is  evident  that  pi}  p.,,  •••,  pn  are  also  symmetric  functions 
in  12  of  «!,  a2,  •••,  an  ;  for,  an  interchange  of,  say  «  and  al}  brings 
about  simply  an  interchange  of  N  and  NI>  Since  the  inter- 
change of  N  and  Nt  does  not  alter  these  functions,  the  inter- 
change of  a  and  «!  does  not. 

Now  «i,  «2,  •••,  «„  are  the  roots  of  the  equation  /(#)=0;  hence 
the  coefficients  />15  p2,  •••,  pn  of  $(?/)  =  0,  being  symmetric  func- 
tions in  12  of  a1?  •  ••,  «„,  may  be  expressed  as  functions  in  O  of 
the  coefficients  of/(z)  =  0  (§  70). 

But  by  hypothesis  the  coefficients  of  f(x)  =  0  are  numbers 
belonging  to  the  domain  O,  hence  the  same  thing  is  true  of 
2h,  "-)Pn-  Thus  <£(?/)  =  0  is  an  equation  of  the  nth  degree  in  12, 
having  the  roots  N,  N1}  •  •  •,  Nn_i. 

Ex.  1.  As  an  illustration,  let  /  (x~)  =  x*  +  1  =  0,  then  ft  =  Q(i)  and  the 
roots  are  ±  \  \/2(l  +  f),  ±  \  V2(l  —  i).  If  a  —  %  V2(l  +  z),  the  domain 
ii(it  a)  consists  of  numbers  «  +  ife,  where  a  and  b  may  be  rational,  or 
irrational  involving  V2.  Let  JV  =  a3  +  a~  +  «+  1,  then  .Y  =  1  +  (1  +  V^)i, 
and  the  numbers  conjugate  to  it  are, 

N  =  1  +  (1  +  V2)f,  JV2  =  1  -  (1  +  V2)z, 


and  *(y)  =  (y  -  2V)  (y  -  JVi)  (y  -  JV2)  (2/  -  JV8) 

=  y4  _  4  y3  +  12  y2  _  lg  y  +  g  _  0. 

Thus,  Ar  and  the  numbers  conjugate  to  it  are  roots  of  an  algebraic  equa- 
tion of  the  fourth  degree  in  fia),  that  is,  *(y)  =0  is  an  equation  in  the 
same  domain  as/(x)  =  0,  and  both  are  of  the  same  degree. 

Ex.  2.  Show  that  5,  z,  \/2  are  each  numbers  lying  in  the  domain  n(a) 
of  Ex.  1,  and  that  each  is  a  root  of  some  reducible  equation  of  the 
fourth  degree. 

137.  Theorem.  Every  number  of  the  domain  n(o)  can  be 
expressed  as  a  function  in  O  of  any  primitive  number  N  of  the 
domain  O(a). 

Let  N'  be  any  number  in  O(a)  and  N1,  N\,  N'2,  •••,  N'n_l  the 
numbers  conjugate  to  it.  Let 

*(a?)  =  (*  -  N)(x  -  N,)  -  (x  -  JVn_0, 


GALOIS   THEORY   OF   ALGEBRAIC    NUMBERS          147 

where  JV,  N1}  •••  Nn^  are  conjugates  to  'the  primitive  number 
JV".     We  now  construct  a  new  function,  ^(03),  as  follows  : 


,         -i 

X         JV  n_j 

This  is  a  function  of  x  of  the  (n  —  l)th  degree, 
Since  N  =</>(«),          -JVi  =0(a1),  ••-, 

and  JV'  =  &(«),          JV'^^i),-", 

it  follows  that  an  interchange  of,  say,  «  and  «!  interchanges  not 
only  JV  and  JV],  but  also  N'  and  JV'j,  and  also  the  first  two 
fractions  in  the  expression  for  ^(»). 

But  $(a?)  is  not  affected  by  such  an  interchange.  Hence  \j/(x) 
is  not  affected,  no  matter  what  two  a's  replace  each  other. 
From  this  it  follows  that  \j/(x)  is  a  symmetric  function  of 
a,  «i,  •••,  an_l  in  O  and  the  coefficients  of  \j/(x)  are  numbers  in  O. 

If  now  we  put  x  =  JV,  then  <I>(JV)  =  0.  As  N  is  primitive 
and  consequently  different  from  JVlt  JV2,  •  ••,  it  follows  that  each 
fraction  in  ty(x)}  except  the  first,  is  zero  when  x  =  JV;  for,  it  has 
a  numerator  that  is  zero  and  a  denominator  that  is  finite. 

The  first  fraction  gives  us  -.     By  §  20  we  have,  for  this  inde- 

terminate, the  relation  ----  ^  —  *-  =  JV'4>'(JV),  where  <$'  means  the 

differential  coefficient  of  <J>  with  respect  to  x.  This  relation 
yields  «/<JV)  =  N'&(N)  or  JV'  =  i^(JV)/*'(JV),  where  $'(JV)  is 
not  zero,  because  <£(&•)  has  no  multiple  roots.  Since  i/^JV)  and 
O'(JV)  are  both  functions  of  Niu  O,  it  follows  that  any  number 
2V'  can  be  expressed  as  a  function  in  O  of  any  primitive 
number  JV. 

Ex.  1.  Prove  that  the  domain  i2(Ar)  is  identical  with  the  domain  O(a), 
N  being  primitive  in  O(a). 

Ex.  2.    It  was  shown  in  Ex.  2,  §  135,  that  N  =  -  ~  is  a  primitive 

V2 

number  of  12(1,^)  >  where  ±  V2  are  the  roots  of  the  irreducible  equation 
a;2  —  2  =  0.  Express  5  +  3  V2,  5  and  \/2,  as  functions  of  N  in  I2(i). 

Ex.  3.   Express  5,  i,  V%  in  Ex.  2,  §  136,  each  as  a  function  in  ft  of  a. 


148  THEORY    OF   EQUATIONS 

138.  Theorem.  If  N  is  primitive  in  fi(tt),  then  the  numbers 
N,  ND  •  •',  Nn_!  are  roots  of  an  irreducible  equation  $>(x)  =  0  of 
the  nth  degree;  if  N  is  imprimitive,  then  these  numbers  may  be 
divided  into  nt  sets  of  n.2  equal  numbers  in  each  set,  and  3>(x)  =  0 
is  the  njh  power  of  an  irreducible  equation  of  the  njth  degree. 

If  *(a?)  =  (a;  -  N)(x  -  NJ  •••(«-  Nn_,)  =  0 

is  reducible,  decompose  it  into  its  irreducible  factors.  Take  one 
of  these  irreducible  factors,  say  Q(x).  Then  0(x)  —  0  must 
have  as  a  root  at  least  one  of  the  numbers  N*  Nl}  •  ••,  ^n_j. 
Let  JVi  be  such  a  root.  Then  6(N1)  =  0,  and  since  ^Vj  =  <£(«i), 
§  134,  we  have  0[<£(ai)]  =  0 ;  that  is,  0[<K-T)]  =  0  has  «j  for  one 
of  its  roots.  Thus  0[<£(X)]  =  0  and/(V)  =  0  are  two  algebraic 
equations  having  a  common  root,  namely  a^  As  f(x)  =  0  was 
assumed  to  be  irreducible,  it  follows  by  §  126  that  each  of  the 
roots  a,  au  •••,  «„_!  of  the  equation  f(x)  =  0  must  satisfy 
0[</>(o;)]  =  0.  Remembering  that  N{  =  <£(«<),  we  see  that  each 
of  the  numbers  N,  JV^  •••,  Nn_t  must  satisfy  the  equation 
B(x)  =  0. 

Now  if  N,  N1}  •••,  ^Vn_i  are  all  distinct,  then  0(x)  —  0  must 
be  of  the  nth  degree,  and  <£(#)  =  0  and  6(x)  =  0  are  identical ; 
since,  by  hypothesis,  0(x)  =  0  is  irreducible,  3>(#)  =  0  must  be 
irreducible. 

If,  on  the  other  hand,  some  of  the  roots  ^7",  N1}  •••,  Nn^  are 
alike;  let  N,  Nl}  •••,  N^_i  represent  the  distinct  roots,  then  the 
irreducible  equation  0(x)  =  0  is  of  the  degree  n^  Any  other 
irreducible  equation,  ^(z)  =  0,  obtained  by  factoring  <J>(#)  =  0, 
must  be  satisfied  by  at  least  one  of  the  set  of  roots  N,  N^  •  •  •,  yni_i, 
for,  every  multiple  root  in  <J>(.r)  =0  has  one  representative  in  the 
list  of  distinct  roots ;  hence  O^x)  =  0  must  be  satisfied  by  each 
loot  in  the  set  and  is  identical  with  the  equation  0(x)  =0,  the 
two  having  all  their  nl  roots  in  common. 

It  thus  appears  that  if  <£(#)  =  0  is  reducible  and  is  resolved 
into  its  irreducible  factors,  these  factors  are  identical  to  each 
other.  Thus,  &(x)  =  0  is  a  power  of  0(x)  =  0.  Since  <J>(a;)  =  0 


GALOIS   THEORY   OF    ALGEBRAIC    NUMBERS          149 

is  of  the  nth  degree  and  6(x)  =  0  of  the  ?ijth  degree,  n  must  be 
a  multiple  of  n^  that  is,  n  =  n^t.^ 

Ex.  1.  As  an  illustration,  take  the  irreducible  equation  f(x )=x*+ 1=0. 
It  has  the  roots  a  -  |\/2(1  +  z),  «!  =  -  J \/2(l  +  i),  «2  =  +  £V2(1  -  i), 
«3  =  —  \  \/2(l  —  i).  Let  JV=  </>(«)=«-,  then  JV=  JV1=  i  and  A72  =  Na  =  —  i. 
Hence,  <t>(x)  =  (x  +  i)'2(a:  -  *)2  =  (2-  +  I)2  =  0.  We  have  0(x)=x*  +1=0, 
whichis satisfied  by  N,Ni,Nz,Ns.  The  equation  0[>(z)]=0(x2)  =  (z2)2+ 1=0 
is  satisfied  by  «,  «i,  «2,  «3,  the  roots  of /(*)  =  0. 

Ex.  2.  From  the  roots  of  the  equation  in  Ex.  5,  §  133,  find  JV",  JV"i,  JV2,  JV8, 
when  N=a?  +  a3.  Determine  whether  the  equation  <£(a;)  =  0  is  in  this 
case  reducible ;  if  it  is,  find  n\  and  w2  and  show  that  0[<£(a)]  =  0  is  satis- 
fied by  the  roots  of  the  given  equation  /(x)  =  0. 

Ex.  3.  From  the  roots  of  the  equation  in  Ex.  5,  §  133,  find  NI,  N2,  JVj, 
when  N  =  4  a.  Is  4>(x)  =  0  reducible  ? 

Ex.  4.  In  Ex.  5,  §  135,  form  $(?/)  =  0  and  examine  its  reducibility, 
when  N  =  a2. 

139.  Normal  Equations.  A  normal  equation  is  an  irreducible 
equation  in  which  each  root  can  be  expressed  as  a  function  in 
O  of  one  of  the  roots. 

Ex.  1.  The  roots  «i,  «2,  «s,  of  x4  +  1  =  0,  Ex.  1,  §  138,  may  be  expressed 
in  terms  of  a  thus :  <*i  =  —  a,  «2  =  —  a3,  «3  =  +  a3.  Hence  x4  +  1  =  0, 
being  irreducible,  is  normal. 

Ex.  2.   Show  that  x*  +  x3  +  xz  +  x  +  1  =  0  is  a  normal  equation.    <„ 
Ex.  3.    Show  that  x*-2a;2  +  9  =  0is  normal.   v 


CHAPTER   XIV 

NORMAL  DOMAINS 

140.  Theorem.     A  primitive  number  of  a  normal  domain  of 
the  nth  degree  is  a  root  of  a  normal  equation  of  the  nth  degree. 

If  a  number  p  be  adjoined  to  fi,  making  ti(p)  a  domain  of  the 
nth  degree,  every  number  N  in  the  domain  £}(p)  is  a  root  of  an 
equation  F(x)  =  0  of  the  nth  degree  in  Q,  the  other  roots  of 
which  are,  by  §  136,  the  remaining  numbers  conjugate  to  Nt  viz. 

N,,  N»  -,  N^. 

Since  N  is  assumed  to  be  primitive,  F(x)  =  0  is  irreducible 
(§  138). 

Any  number  N()  being  defined  by  <f>(pt),  belongs  to  the  domain 
fi(Pi).  Since  O(p)  is  normal,  we  have  O(p)  =  n(p  }  =  •••  =  fi(Pn_1) 
(§  132).  Hence  all  the  numbers  N,  N^  •••,  Nn_j  belong  to  the 
domain  n(p),  and  can  be  expressed  as  functions  in  Q  of  the 
primitive  number  N  (§  137).  From  this  it  follows  that  F(x)  =  0 
is  a  normal  equation. 

141.  Theorem.      Conversely,  if  p  is  a  root  of  a  normal  equa- 
tion, then  O(p)  is  a  normal  domain  of  the  same  degree  as  that  of 
the  equation. 

Let  po  be  the  root,  of  which  the  other  roots  are  functions  in 
fi;  that  is,  let  pv  —  <f>v(po),  where  v  may  be  1,  2,  •••,  or  (n  —  1). 
Since  p0  is  a  root  of  the  given  irreducible  equation  of  the  «th 
degree,  the  domain  Q(p ,  and  all  the  domains  conjugate  to  it  are 
of  the  nth  degree  (§  132). 

150 


NORMAL   DOMAINS  151 

Any  number  in  the  domain  Q(P>J,  i.e.  in  the  domain  O[(<)  (p  Yj, 
is  a  function  in  O  of  [<#V(PO)]>  and,  therefore,  also  a  function  in 
O  of  po  itself;  that  is,  any  number  in  the  domain  fi[<t  (p  ,-,  occurs 
also  in  O(p  }.  The  converse  is  true  also.  Hence  the  conjugate 
domains  are  identical,  and  !2(p  }  is  a  normal  domain. 

COROLLARY.  Since  the  domain  O^  (p  }]  contains  all  the  roots 
of  the  given  normal  equation,  each  of  these  roots  can  be  ex- 
pressed as  a  function  in  O  of  the  root  </>,,(po)>  where  <£,,(po)  m&y 
represent  any  one  of  the  roots.  Thus,  in  a  normal  equation  every 
root  can  be  expressed  not  only  as  a  function  in  O  o/some  one  root, 
but  as  a  function  in  O  of  any  one  of  the  roots. 

yl    _     1 

Ex.  1.    Show  that  the  equation  -  -  =  0  is  normal. 

x  —  1 

Ex.  2.   Show  that  y*  +  10  y?  +  40  x  +  205  =  0  is  normal. 

142.  Adjunction  of  Several  Magnitudes.  The  adjunction  of 
several  magnitudes  may  be  replaced  by  the  adjunction  of  a  single 
magnitude. 

Let  a,  ft,  y,  •••  be  numbers  adjoined  to  the  domain  O,  giving 
the  enlarged  domain  OUi  J3i  ^  ...).  To  prove  that  a  number  p  can 
be  found,  such  that  the  domains  O(a,  ^  ..0  and  fi(p)  are  identical. 

Let  «  be  one  of  the  roots  a,  «j,  •••,  «m_j  of  an  algebraic  equa- 
tion in  O,  fi(x)  =  0.  Similarly,  let  ft  be  one  of  the  roots 
ft,  Pi,  •">  Pn-i  of  /2(«)  =  0,  y  one  of  the  roots  y,  y1?  y2,  •»,  y0_j  of 
/3(a;)  =  0,  and  so  on.  Without  loss  of  generality  we  may 
assume  that  none  of  these  equations  have  multiple  roots. 
Now  assume  for  p  the  following  linear  function  of  «,  ft,  y,  •••, 


where  a,  6,  c  are  indeterminate  coefficients  to  which  in  special 
cases  any  convenient  numerical  value  in  O  may  be  assigned. 
It  is  evident  that  o  is  a  magnitude  in  O(ai  fr  y>  ...,,  for  it  is  a 
rational  function  of  «,  ft,  y,  •••.  The  expression  for  p  involves 
one  root  from  each  of  the  equations  fi(x)  =  0,  fz(x)  =  0,  •••. 


152  THEORY   OF   EQUATIONS 

Next,  replace  the  roots  a,  /?,  y,  •••  by  any  other  combination 
«!,  fti,  •/!,  •••  of  the  roots,  one  root  being  taken  from  each  equa- 

tion.    We  get 

Pl  =  aal  +  bj3l  +  cy1  +  .-. 

Similarly  we  obtain  p2,  ps,  •••.  The  total  number  of  p's  is 
equal  to  the  total  number  of  possible  combinations,  which  is 
m  •  n  •  o  •••,  where  m,  n,  o  are  respectively  the  degrees  of  the 
equations.  By  assigning  appropriate  values  to  a,  b,  c,  ••-,  all 
the  p's  will  be  distinct  from  each  other. 
Now  construct  the  function  F(t),  thus  : 

F($s  (*-,)(*-*)(«  -*)....  II 

F(f)  is  not  altered  if  «  is  replaced  by  «„  or  /3  by  /?,.  Hence 
l,he  coefficients  of  'II,  obtained  by  performing  the  indicated 
multiplications,  are  symmetric  functions  of  the  roots  of  each 
one  of  the  equations  fi(x)  =  0,  fz(x)  =  0,  •••;  therefore,  the 
coefficients  are  numbers  in  O,  and  F(£)  is  a  function  in  O. 

Now,  any  number  N  in  O(ai  /Ji  yi  ..0  is  a  rational  function  of 
«,  ft,  y,  •••.  Let  -iVgo  over  into  JV,,  N2,  •••  for  the  substitutions 
which  convert  p  into  plt  p2,  •••.  With  these  construct  the  new 
function  G(£),  defined  as  follows  : 

JL  +  J?L+J?L+....          m 


G(f)  is  symmetrical  with  respect  to  the  roots  of  fi(x)  =  0, 
/.,(#)  =  0,  •••.  Hence  its  coefficients  lie  in  O.  For  t=p,  F(t) 
vanishes,  as  appears  from  II.  But  the  denominator  t  —  p  van- 
ishes also. 

Hence  for  t  =  p,  we  have  by  §  20 


p  —  p 
where  F1^)  is  the  first  differential  coefficient  of  F(f). 

Hence,  J 


NORMAL   DOMAINS  153 

This  means  that  N  is  a  rational  function  of  p ;  that  is,  any 
number  in  O(ai  /3)  ...)  is  a  rational  function  of  p,  and  lies,  there- 
fore, in  the  domain  ii(p).  Conversely,  any  number  in  O(p)  lies 
in  O(ai  a  ...)  since  every  number  in  fi(p)  is  a  rational  function  of 
p,  and,  therefore,  of  a,  (3,  y,  ••-.  This  shows  that  O(p)  and 
O(o  p  }  are  coextensive  domains,  and  the  adjunction  of  a,  (3, 
y,  •••  to  O  may  be  replaced  by  the  adjunction  of  p. 

Ex.  1.    Go  over  the  above  proof  for  the  special  case  where 

a  =  \/2,  /3  =  #6,  7  =  5  =  ...  =  0,  a  =  b  =  1,  N  =  3  V2  \/5. 
Here  /i(x)  =  x2  -  2  =  0,   /2(x)  =  x3  -  5  =  0.      Then    p  =  V2  +  \/5. 
There  are  six  different  p's,  and  II  is  of  the  sixth  degree  in  t.     Of  what 
degree  is  III  ? 

G(C)=N(t-piXt-P*)  •••(«- PS)  +  -ffi(*-p)(«-p2)...  («-p6)  +  ••• 
0(p)  =  jy(p  -  pi)(p-P2)  -  (P  -  PS)  =  54°P2  +  360>  where 
p  =  V2  +  \/5,  ps  =  -  V2  +  v^5, 

P!  =  \/2  -f  <i>  v'F,  p4  =  —  V2  +  to  V5, 

P2  =  V2  +  w2  \/5,  p5  =  -  V2  +  «2  v^. 

By  Ex.  14,  §  71,  the  equation  whose  roots  are  p,  pi,  •••,  ps,  is 

F(f)  -  fl  -  6  t*  -  10  P  +  12  «2  -  60  t  +  17  =  0. 
/.  J?"(p)  =  6  p5  -  24  p8  -  30  p2  +  24  p  -  60. 
We  see  that  £(p)  +  F'(p)  =  2\T. 

Ex.  2.  Is  the  adjunction  of  V^2  to  O(i)  equivalent  to  the  adju&v 
tion  of  i  +  V2  ? 

Ex.  3.  Are  the  two  domains  %,  rf,  ^3)  and  %,  ^  coextensive  ?  I£ 
not,  is  one  a  divisor  of  the  other  ? 

143.  The  Galois  Domain.  If  /(a?)  =  0  is  an  equation  of  the 
7ith  degree  with  distinct  roots  a,  «!,  •••,  an_},  then  the  domain 
Q,  .,  obtained  by  the  adjunction  of  all  its  roots  to  O, 

"  (a,  Oj)  •••  <*n_j"  "* 

is  called  the  Galois  domain  of  the  equation  f(x)  =  0.  Thus 
the  roots  of  the  cubic  tf  +  Sx2 -2z-6  =  0  are  -3,  ±V2; 
hence  its  Galois  domain  is  fyi.^)- 

Ex.  1.    Find  the  Galois  domain  of  x4  +  6  jc2  +  5  =  0. 

Ex.  2.  Find  the  Galois  domain  of  the  equation  in  Ex.  5,  §  133.  Sho* 
that,  in  this  case,  i2(o,  ,H, ...  On_t)  =  O(a)  =  ft(at)  =  n(aa)  =  n(a,V 


154  THEORY   OF   EQUATIONS 

144.    Theorem.     The  Galois  domain  of  any  algebraic  equation 
is  a  normal  domain. 

The  degree  of  the  Galois  domain  O(ai  ai>  ...,  BB_I)  is  not  usually 
the  same  as  that  of  the  equation  f(x)  =  0  ;  let  it  be  m. 
Let  p  be  a  primitive  number  of  the  Galois  domain,  then 


It  follows  that  p  is  a  root  of  an  irreducible  equation  of  the 
degree  m  (§  138),  viz.  the  equation 


The  root  p,  being  a  number  in  the  Galois  domain,  can  be 
expressed  as  a  function  of  OQ,  a1}  •••,  aB_1?  in  J7;  that  is, 


Consider  all  the  permutations  which  can  be  performed  with 
the  n  subscripts  of  the  letters  «,  taken  all  at  a  time.  The 
number  of  these  permutations  is  n  !  They  correspond  to  the 
symmetric  group  of  substitutions  (§  98). 

If  we  operate  upon  the  subscripts  in  II  with  each  substitu- 
tion of  the  symmetric  group  of  the  order  n!,  in  turn,  we  obtain 
values  for  p  which  we  indicate,  respectively,  by 

p>  pi,  •->  Pn\-i-  III 

Next,  if  we  operate  with  any  substitution  of  the  symmetric 
group  upon  the  p's  in  III,  we  get  the  same  set  of  p's  over  again, 
only  in  a  different  order  ;  for,  any  number  resulting  from  this 
second  operation  is  obtained  from  II  by  two  substitutions,  the 
product  of  which,  by  definition  of  a  group,  is  identical  with 
one  of  the  substitutions  in  the  group.  Hence,  if  we  form  the 
equation  H(y}  ^(y_  ^  _  pj)  ...  (2/  _  ^  =  0>  IV 

this  equation  is  invariant  under  any  of  the  substitutions  of 
the  symmetric  group  ;  hence,  the  coefficients  of  y,  obtained  by 
performing  the  indicated  multiplications  in  IV,  are  invariant. 


NORMAL    DOMAINS  155 

But  these  coefficients  are  functions  in  O  of  the  roots  p,  pl}  •  ••, 
and  by  relation  II,  also  functions  in  O  of  «0,  u1}  •••,  «M_I. 

Because  of  the  invariance  of  the  coefficients  of  IV  under  the 
symmetric  group,  they  are  symmetric  functions  in  O  of  «0,  a1} 
'"}  an-i>  i-e-  symmetric  functions  in  O  of  the  roots  of  f(x)  =0. 
Hence  IV  is  an  equation  in  O  (§  123),  and  its  roots  are  numbers 

in  0(«,  ...,»„_!>• 

But  p  is  a  root  of  both  H(y)  =  0  and  g(y)  =  0.  Since  g(y)  =  0 
is  irreducible,  all  its  roots  must  be  roots  of  H(y)  =  0  (§  126). 
But  all  the  roots  of  H(y)  =  0  are  numbers  in  O(ai  m>  a?j  _j)  ; 
hence  all  the  roots  of  g(y)  =  0  (viz.  the  conjugate  numbers 
Pt  Pi,  "•>  Pm-i)  are  numbers  in  O(a>  ...,  Bn_1).  But 


hence  we  have        O(p)  =  O(PI)  =  •  •  •  =  O(Pm_l). 
That  is,  O(a,  ...,an_1  is  a  normal  domain. 

145.  Galois  Resolvent.  The  equation  g(y)  =  0  of  §  144  is 
called  the  Galois  resolvent  of  the  given  equation  f(x)  =  0  in  the 
domain  O,  defined  by  the  coefficients  of  the  equation  f(x)  =  0. 
This  resolvent  possesses  the  following  properties  : 

(1)  g(y)  =  Q  is  irreducible. 

(2)  Each  root  of  f(x)  =  0  can  be  expressed  as  a  function  in  Q 
of  one  root  p  of  the  equation  g(y}  =  0.     That  is,  each  of  the 
roots  a,  «j,  ••-,  «n_j  occurs  in  O(a>  ...,  an_1>,  a  domain  equivalent 
to  O(p). 

(3)  One  root  p  of  g(y)  =  0  can  be  expressed  as  a  function  in  Q 
of  the  n  roots  off(x)  =  0.     That  is,  by  II,  §  144,  we  have 

P=/i(«o>  «i>  —,*n-i)- 

Ex.  1.   The  cubic  x3  +  3  a;2  +  x  -  I  =  0  has  the  roots 
a  =  -  1,  «i  =  -  1  +  V2,  «2  =  -  1  -  V2. 

Hence    the    Galois   domain    is    «(itN/2)-     Also»  P  ="v/2  ™  a  root  of  the 
irreducible  equation  g(y)  =  a;2  -  2  =  0  and  is  a  primitive  number  of  the 


156  THEORY   OF    EQUATIONS 

Galois  domain.  The  equation  x'2  —  2  =  0  is  a  Galois  resolvent,  because 
(1)  it  is  irreducible  ;  (2)  the  root  a  =  —  V2  I  V2  and  the  roots  «i,  «2  are 
each  functions  in  ii(i)  of  V2  ;  (3)  we  may  express  p  as  a  function  of 
a,  «i,  «2>  thus,  p—  V2  =  «22  —  «i  +  4  a. 

Ex.  2.  Show  that  in  Ex.  1,  p  =  a  +  b  V2,  which  is  a  root  of  the  equa- 
tion x2  —  2  aa;  +  a'2  —  2  62  =  0,  a  and  6  being  rational,  is  a  primitive 
number  in  the  domain  0(V/2)»  and  that  this  quadratic  is  a  Galois  resol- 
vent of  the  given  cubic. 

Ex.  3.  Show  that  the  degree  of  the  Galois  resolvent  of  an  equation  of 
the  ?ith  degree  cannot  exceed  n  !.  See  §  144. 

Ex.  4.  Construct  the  equation  H(y)  —  0  of  §  144  for  the  general  cubic 
ce8  +  ai»2  +  a2x  +  0,3  =  0,  whose  roots  are  «,  «i,  «2. 

As  in  §  142  select  appropriate  values  in  O  for  the  coefficients  c,  ci,  C2, 
so  that  distinct  values  for  p  are  obtained  for  every  permutation  of  the 
roots  «,  «i,  «2  in  the  relation  p  =  ca  +  Ci«i  +  c2«2. 

Performing  upon  this  the  six  substitutions  of  the  symmetric  group  of 
the  third  degree,  §  104,  we  obtain 


Pi  =  c«i  +  Ci«2  +  c2«, 


p'2  =  C«2  +  Ci«i  +  C2«. 

We  first  form  the  cubic  -whose  roots  are  p,  pi,  p2.     We  get 


2ppi  =  Sc2  •  2««i  +  2a2  •  Zcci  + 
=  a2Sc2  +  («i2  -  a2)Scci. 

To  obtain  the  product  ppip2,  observe  that  the  terms  cciC2«8,  cciC2«i3, 
cciC2«28  occur  in  the  product  ;  their  sum  is  ccic^^a3.  Since  c,  ci,  c2  and 
a,  «i,  «2  are  similarly  involved,  the  expression  ««i«22c3,  also  occurs  in 
the  product.  The  term  cciC2««i«2  occurs  three  times  ;  hence  we  have 


Observe  that  a2«i  has  in  the  product  the  coefficient  pc=c2Ci  +  Ci2c2  +  c22c 
and  that  «i2«2  and  «22«  have  each  this  same  coefficient.  Hence  pcpa  ig 
part  of  the  product,  where  pa  =  a?ai  +  «i2«2  +  «22«.  Similarly  <x«i2, 
«i«22,  «2«2  have  each  the  coefficient  p'c  =  cci2  +  CiC22  +  CaC2.  Therefore, 


NORMAL    DOMAINS  157 


P'cp'a  occurs  in  the  product,  where  p'a  =  aa^  +  a^ctf  +  e^a2.  We  have 
now  found  all  together  27  terms  which  belong  to  the  product  ppip2  ;  they 
constitute  the  entire  product.  That  is, 


4-  «ai«2Sc3  +  3  cciC2aai02  +  pcpa  +p' 
We  get 

Pa  +  p'a.  =  2«  •  2««i  -  3  ««!«;,  =  3  a3  -  a^a*  =  qa, 

Pa  -P'a  =  ««!(«  -  «l)  +  «22(«  -  «l)  -  «2(«2  -  «!« 

=  (a  -  «i)  (a  -  a2)  («i  —  «•_>)  =  VZ>~, 
where  Z>0  Is  the  discriminant  of  the  given  cubic,  hence 


Similarly  we  have  2pc  —  qc+  VZ>C, 

2p'c  =  ^'c-VA. 
Hence 


-  «i8  -  3  a3)  -  a3Sc3  -  3  cc1c2a8 

We  have  now  found  the  coefficients  of  the  cubic  whose  roots  are  p,  pi,  p2, 
expressed  in  terms  of  the  coefficients  of  the  given  cubic. 

In  finding  the  coefficients  of  the  cubic  whose  roots  are  pf,  p'1?  p'2  we 
notice  that  2p'  =  S/»,  and  Sp'p'i  =  Sppi-  The  product  p'/o'ip'a  differs  from 
ppxp2  only  in  the  sign  before  the  radical.  Consequently,  on  multiplying 
the  left  members  of  the  two  cubics,  the  radical  disappears  and  we  obtain 
a  sextic,  whose  coefficients  are  numbers  in  ft.  This  sextic  is  the  required 
equation  H(y)  =  0,  whose  roots  are  p,  pi,  p2,  p',  p'i,  p'2- 

Ex.  5.  Show  that  when  in  the  sextic  of  Ex.  4  the  value  of  Da  is  a  per 
feet  square,  the  sextic  becomes  reducible  into  two  cubic  equations  in  ti. 
Hence  <?(?/)  =  0  is  a  cubic  in  this  instance. 

Ex.  6.   Of  what  degree  is  the  Galois  resolvent  of  the  general  quartic  ?    - 
The  general  quintic  ? 

Ex.  7.  Find  the  roots  of  the  equation  x5  +  je*  —  x3  —  a;2  —  2  x  —  2  =  0. 
From  the  roots  determine  the  Galois  domain.  Prove  that  x4  —  2z2  +  9  =  0 
is  a  Galois  resolvent. 

146.  Theorem.  The  Galois  resolvent  is  a  normal  equation, 
and  any  normal  equation  is  its  own  Galois  resolvent. 

The  resolvent  is  a  normal  equation  because  (1)  it  is  irre- 
ducible and  (2)  all  its  roots  occur  in  the  Galois  .  domain 


158  THEORY   OF   EQUATIONS 

where  p  is  a  root  of  the  resolvent  (§  144),  and  are,  therefore, 
functions  in  O  of  the  one  root  p  (§  138). 

To  prove  the  second  part,  let  f(x)  =  0  be  a  normal  equation, 
having  the  roots  a,  a1}  •••,  «n_j.  Then  O(a)  is  a  normal  domain 
(§  141) ;  /(a?)  =  0  is  its  own  Galois  resolvent,  because  being 
irreducible  it  satisfies  property  (1)  in  §  145,  and  all  its  roots 
being  in  the  domain  O(a),  and,  therefore,  functions  of  a  in  O,  it 
satisfies  also  properties  (2)  and  (3). 

Ex.  1.  Show  that  the  equation  in  Ex.  5  (§  133)  is  its  own  Galois 
resolvent. 

Ex.  2.  Show  that  the  Galois  resolvent  in  Ex.  2  (§  145)  satisfies  the 
definition  of  a  normal  equation. 

Ex.  3.  Find  the  Galois  domain  for  the  equation  in  Ex.  3  (§  133). 
Find  the  irreducible  equation  in  Qm  having  the  primitive  number  Vd  +  V5 
as  a  root.  Show  that  this  equation  is  its  own  Galois  resolvent  and  that 
the  Galois  domain  is  normal. 

147.  Theorem.  If  f(x)  =  0  is  a  normal  equation  of  the  nth 
degree  with  a  root  p  as  a  primitive  number  in  the  normal  domain 
O(P),  then  the  transposition  (ppA)  causes  each  of  the  numbers 
conjugate  to  p  to  be  replaced  by  some  other  of  their  own  set,  but  no 
two  numbers  are  replaced  by  the  same  one. 

Let  the  numbers  conjugate  to  p  be  p,  Pl,  •  ••,  pn_lf  They  are 
all  roots  of  the  equation  f(x)  =  0  (§  138).  Since  O(p)  is 
assumed  to  be  normal,  they  are  contained  in  it.  Hence  we 

P  =  &G»)j  Pl  =  <£l(p)>    "•)  Pn-l  =  <£n-l(p)> 

where  <£„,  fa,  •••  are  functions  in  O.  If  in  <£A(/o),  which  is  a  root 
of  f(x)  =  0,  we  replace  p  by  ph,  we  get  as  a  result  <£A(pA),  which, 
being  conjugate  to  <f>k(p),  is  another  root  of  f(x)  =  0  (§  136). 
Hence  the  numbers  in  the  series 

^»O(PA)J  <£i(p»)j  '•')  <k»-i(/>») 

are  identical  with  numbers  in  I,  except  in  the  order  in  which 
they  are  written.  Now,  if  we  can  show  that  the  roots  II  are 
all  distinct,  our  theorem  is  proved. 


NOKMAL  DOMAINS  159 

None  of  the  roots  II  are  alike,  for  suppose  <&(/3A)  =  <£*(pA)> 


thatis'  -  in 


then  III  is  an  equation  having  ph  as  a  root.  But  the  irreducible 
equation  f(x)  =  0  has  also  ph  as  a  root.  Hence  III  must  be 
satisfied  by  all  the  roots  of  f(x)  =  0  ;  for  instance,  by  p.  Con- 
sequeutly,  _  =  0. 


This  equation  by  I  may  be  written  p{  —  pk  =  0,  which  cannot  be 
true,  since  p  is  a  primitive  number. 

Ex.  1.  In  Ex.  5,  §  133,  we  have  given  an  irreducible  equation  with  the 
roots  p,  pi,  pg,  ps,  conjugate  to  p  in  the  normal  domain  O(P).  We  have 
Pi  =  p2,  pa  —  P3>  ps  =  p4-  Hence  the  roots  may  be  represented  by  the 
series  00,1  T 

p,  p2,  p3,  p4.  i 

If  in  I  we  write  p3  for  p,  we  get 

PS,  ps2,  ps8,  Pa4, 

where  ps2  =  p2,   pa8  =  pi,  ps4  =  p.      Hence  the  transposition   (pps)   only 
changed  the  order  of  the  roots. 

Ex.  2.  What  is  the  order  of  the  roots,  if  in  Ex.  1,  we  apply  the 
transposition  (pp2)  ? 

148.  Theorem.  Every  transposition  (phpk)  in  the  normal 
domain  O(p)  is  equal  to  some  one  of  the  transpositions  (ppi)> 


We  have  pft=<^A(/o),  I 

where  <£ft(p)  is  a  root  of  the  normal  equation  f(x)  =  0.  Upon 
<£A(p)  perform  the  transposition  (ppt),  and  we  get  <£ft(/o,-).  This 
is  a  number  conjugate  to  <£A(/o),  and  is,  therefore,  one  of  the 
other  roots  of  f(x)  =  0,  say  pk  (§  138),  so  that 


Since  the  transposition  (pAp*)  changes  ph  to  pk,  and  the  trans- 
position (ppf)  changes  <K(p)  to  ^>A(/3,),  we  have  from  equations 
I  and  II  that  (p*pk)  = 


160  THEORY   OF    EQUATIONS 

Ex.  1.    In  Ex.  5,  §  133,  the  four  roots  satisfy  the  following  relations : 


P  =P2, 
P2  =  P24, 
Ps  =  Pz, 
P*  =  P-/. 


Operate  upon  the  left  members  of  these  equalities  with  the  transposition 
(pp2) ,  and  upon  the  right  members  with  (psps),  and  show  that  (p/>2)  =  (pzps)- 
Ex.  2.    In  Ex.  1  find  the  transposition  (p/>j)  which  is  equal  to  (pips). 
Ex.  3.    In  Ex.  1,  §  136,  find  i  so  that  (««,-)  =  («i«2). 

149.  Substitutions  of  the  Domain  Q(p).  Since  any  transposi- 
tion (phpk)  =  (ppi),  where  i  is  some  one  of  the  numbers  0,  1,  2, 
•••  (n  —  1),  it  follows  that  there  are  not  more  than  n  distinct 
transpositions  in  the  given  normal  domain  O(p),  which  number 
agrees  with  the  degree  of  the  domain  and  the  degree  of  the 
equation  f(x)  =  0,  whose  roots  define  this  domain.  Since  every 
number  in  O(p)  can  be  expressed  as  a  function  of  p  in  O,  since 
every  number  operated  upon  by  (pp,)  passes  into  some  other 
number  in  the  domain  conjugate  to  it,  since,  moreover,  no  two 
numbers  pass  into  the  same  number  (§  147),  it  follows  that 
each  such  substitution  applied  to  all  the  numbers  in  the  normal 
domain  leaves  the  domain  as  a  whole  unchanged. 

The  substitutions  (ppf),  where  i  takes  successively  the  values 
0,  1,  •••  (n  —  1),  are  called  the  substitutions  of  the  domain  O(p). 

If  N=  <f>(p~)  is  invariant  under  (pp<)  so  that  N=  <^(p)  =  ^>(/aj), 
then  we  say  that  N  admits  of  the  substitution  (pp{}.  Observe 
the  difference  between  the  expressions  admits  and  belongs  to 
(§  111).  In  both  the  function  must  be  unaltered  under  the 
substitutions  of  a  certain  group  G^  but  in  the  latter  expression 
we  have  the  additional  condition  that  the  function  must  be 
altered  by  every  substitution  of  G  which  does  not  occur  in  G^ 
GI  being  regarded  as  a  sub-group  of  G. 

If  N=  <f>(p~)  is  a  primitive  number,  then  it  is  distinct  from 
each  of  its  other  conjugates  <£(/°i),  <Kp2)>  •">  <Kpn-i)-  Hence  N 
admits  of  none  of  the  substitutions  (/3pf),  except,  of  course,  the 
identical  substitution  1. 


NORMAL  DOMAINS  161 

150.  Theorem.      The  substitutions  of  the  normal  domain  Q(p) 
constitute  a  group  of  the  order  n. 

Eemembering  the  definition  of  a  substitution  group  (§  95), 
we  need  only  show  that  in  the  n  distinct  transpositions  the 
product  of  any  two,  say  of  (ppi)  and  (pph),  is  equal  to  some  one 
of  the  transpositions  in  the  set,  say  (ppk). 

By  §  148  we  know  that  (ppi)  =  (p^).  Multiply  both  sides 
bJ  (pp»)>  and  we  get 

(PP*)  (fPi}  =  G>/>»)  (W*)  =  (ppk)  5 

that  is,  the  product  of  any  two  substitutions  (ppA)  and  (>/>,)  is  a 
substitution  belonging  to  the  set. 

151.  Theorem.     If  the  equation  f(x)  =  0    yields  the  Galois 
domain  O(p),  then  there  corresponds  to  the  group  of  substitutions 
(fipi)  °f  that  domain  a  group  of  substitutions  s{  of  the  same  order 
among  the  roots  of  the  equation,  such  that  the  product  of  any  two 
substitutions  (PPt),  (PP})  of  the  domain  corresponds  to  the  product 
of  the  two  corresponding  substitutions  si}  Sj  of  the  roots  off(x)=0. 

Let/(#)=0  have  the  roots  a,  a1}  •••,  an_l}  all  of  them  dis- 
tinct. Since  these  roots  are  numbers  in  the  Galois  domain 
O0i  ...  an  =  O(P)  of  the  degree  m,  it  follows  that 


and  that  a,  =  <#>,(p)  where  s  has  any  value  0,  1,  •••  (n  —  1). 
Substituting  for  the  n's  their  values,  we  get  from  I, 


Now  p  is  a  primitive  number  in  the  Galois  domain  O(P)  (§  144), 
and  is,  therefore,  a  root  of  the  Galois  resolvent  g(y)  =  0,  whose 
other  roots  are  the  remaining  numbers  conjugate  to  it,  viz. 
Pu  •••>  Pm-i-  Consider  II  an  equation  having  a  root  P,  then  the 
irreducible  equation  g(y)  =  0  and  the  equation  II  have  p  as  a 
common  root  ;  hence  the  conjugates  of  p  are  roots  common  to 


162  THEORY   OF    EQUATIONS 

both  equations  (§  126).     Eeplacing  p  by  any  of  its  conjugates 
pi,  we  have,  therefore, 


Eeplacing  in  II  p  by  pfl  where  i  and  j  are  distinct,  we  get 

IV 


Since  a,  is  a  root  of  f(x)  =  0  and  a,  =  <£.(p),  we  have  the 
equation  /[<£,(p)]  =0,  which  has  p  as  one  of  its  roots.  But  p 
is  also  a  root  of  the  irreducible  equation  gr(y)  =  0;  hence 
(§  126)  we  have  /[<£.(p,-)]  =  0  ;  that  is,  <k(p,-)  is  some  one  of  the 
roots  «<  of  the  equation  f(x)  =  0.  For  the  same  reason  <p,(pj)  is 
some  one  of  these  roots. 

Since  <j>,(p{)  and  <£,(/>,•)  represent  each  some  root  of  /(#)  =  0,  we 
see  that  in  each  bracket  of  III  and  IV  we  have  some  arrange- 
ment of  the  roots  «,  al}  •••,  an_1. 

The  two  arrangements  are  not  identical  ;  for  if  they  were, 
we  would  have  <£s(pi)  =  <£,(/>;)  for  all  values  of  s  ;  the  right 
members  of  III  and  IV  being  equal,  the  left  members  would 
be  ;  that  is,  pt  =  PJ.  But  this  is  impossible,  since  they  are  roots 
of  the  irreducible  equation  g(y)  =  0,  and  can,  therefore,  not  be 
equal.  Hence  it  follows  that  to  any  two  distinct  substitutions 
(ppt),(ppj)  there  correspond  two  distinct  substitutions  among  the  a?s. 

From  this  we  draw  the  further  conclusion  that  since  the  a's 
belong  to  the  domain  fl(p),  and  the  entire  domain  has  only  m 
distinct  substitutions,  there  are  just  m  distinct  substitutions 
among  the  a's.  There  exists,  therefore,  a  one-to-one  corre- 
spondence between  the  substitutions  (/op,-)  and  the  substitutions 
st  of  the  roots  a. 

Now  the  product  (p/>,-)  (pp^-)  is  equal  to  some  other  substitu- 
tion in  the  group,  say  (pp4).  If  to  (pp4),  (pp,-),  (ppt)  there  corre- 
spond, respectively,  si}  sj}  sk  among  the  roots,  and  if 


we  have  also  st  •  s,  =  sk 


NORMAL   DOMAINS  163 

Ex.  1.   The  quartic  equation  x*  -  12  x3  +  12  x2  +  176  x  -  96  =  0  has 
the  roots 


«3  =  4-2\/3. 

The  Galois  domain  0(p)  is  obtained  by  adjoining  \/7  +  V3  to  fim.     We 
have 


p2  =  -  V?  +  V3,  ps  =  -  \/7  -  V3. 

By  inspection,  we  get 

«  =000  =  2 


=  4  +  1(16  p  -  p3). 

Substituting  for  p,  in  succession,  p,  pl5  p2,  p3,  we  obtain  the  following 
table  • 

"    +00    =«l  ^iC/3)    =«!»  02(p)    =«2,  03(p)    =«g.  I 


a3.  Ill 

«2.  IV 

Operating  upon  ^(p),  0i(p),  02(p),  0s(p)  in  line  I  with  the  transposi- 
tion (ppi)  gives  us  line  II.  The  arrangement  a,  al5  «2,  «3  in  line  I  has 
changed  to  the  arrangement  a,  «i,  «3,  «2  in  line  II.  Hence  (ppi)  corre- 
sponds to  («2  «3).  Thus,  to  the  substitutions  of  the  domain,  viz., 

1»     0>Pi)i     (PP2),     (pps),  V 

there  correspond,  respectively,  the  substitutions  among  the  roots 

VI 


The  latter  are  readily  seen  to  constitute  a  group.  Groups  related  to 
each  other,  as  are  these  two,  are  called  isomorphic.  Group  VI  is  called 
the  Galois  group  of  the  given  quartic  equation. 

Ex.  2.  Find  in  the  list  of  groups  enumerated  in  §  104  the  group  VI  of 
Ex.  1. 


Ex.  3.    In  Ex.  1,  020)  =  «2  and 
Show  that,  in  the  set  of  substitutions  V,  (ppi)(pp2)  =  (pp3).     Forming  all 
possible  products  of  two  transpositions,  show  that  V  is  actually  a  group. 

Ex.4.  The  cubic  x3  +  3x2  +  x  -  1  =  0  has  the  roots  «  =  -!, 
«i  =  —  1+V2,  «2=—  1—  V2  and  the  Galois  domain  12(1,  ••?>,  where 
p  =  V2  and  pi  =  —  v/2.  Find  the  Galois  group  in  both  forms. 


164  THEORY    OF    EQUATIONS 

152.  Galois  Group  of  f(x}  =  0  in  ft.  The  group  of  substitu- 
tions among  the  roots  a,  a^  •  ••,  an^  of  the  equation  f(x)  =  0 
corresponding  to  (isomorphic  with)  the  group  of  the  Galois 
domain  fi(p)  of  that  equation  is  called  the  Galois  group  of  the 
equation.  The  term  Galois  group  is  really  applicable  to  the 
two  isomorphic  groups  indifferently.  For  two  (simply)  isomor- 
phic groups  are  identical,  abstractly  considered,  since  to  every 
substitution  of  one  there  corresponds  a  single  substitution  of 
the  other,  and  vice  versa,  and  since  to  the  product  of  any  two 
substitutions  in  the  one  there  corresponds  the  product  of  the 
two  corresponding  substitutions  in  the  other.  For  convenience 
we  shall  restrict  the  term  Galois  group  to  the  group  of  substi- 
tutions having  the  roots  as  elements. 


Ex.  1.    Show  that  G^4)  and  6r2(2)  are  isomorphic  ;  also  6?6(5)I  and  G& 
Ex.  2.    Show  that  (re(3)  is  simply  isomorphic  with 


153.  Theorem.  Every  function  in  li,  /(a,  aa,  •••,  an_i),  which 
equals  a  number  N  in  O,  admits  every  substitution  of  the  Galois 
group  of  f(x)  =  0. 

Since  O(a,  ^  ....  ^.D  =  O(p),  each  au  where  i  =  0,  1,  •••,(»  —  1), 
is  a  function  in  fi  of  p.  Hence  we  have 

f(a,  «!,-,«„_,)  =  0(p)  =  X,  I 

where  /  and  6  are  functions  in  O.  We  have  0(p)  —  N=  0,  and 
this  equation  in  O  is  satisfied  by  one  root  p,  and  therefore  by 
all  the  roots  which  belong  to  the  Galois  resolvent  g(y~)  =  0 
(§  126).  That  is,  0(p<)  =  N.  But  by  I  the  transposition  (pp.), 
performed  upon  0(j>),  produces  the  same  result  as  the  corre- 
sponding substitution  of  the  Galois  group,  performed  upon 
/(«,  •••,  «n_i).  As  0(p)  remains  unaltered,  so  /(a,  •••,  an_i) 
remains  unaltered. 


NORMAL   DOMAINS  165 

154.  Theorem.     Every  function  in  12,  /(«,  «,,  ...,  an_^),  which 
admits  all  the  substitutions  of  the  Galois  group,  is  a  number  in  12. 

In  the  equation  /(«,  al9  -•-,  aB_a)  =  0(p), 

given  in  §  153,  /(a,  «j,  •••,  «„_!)  admits  by  hypothesis  of  the 
substitutions  of  the  Galois  group;  consequently,  B(p)  admits 
of  the  corresponding  transpositions  of  the  Galois  domain  ll(p). 
That  is,  6(jp),  being  invariant,  is  equal  to  all  its  conjugates  0(p,). 

But  0(p)  is  a  number  in  the  domain  12(p)  and  is  a  root  of  an 
equation  of  the  nth  degree  in  O,  whose  other  roots  are  the 
remaining  numbers  conjugate  to  it  (§  136).  All  these  roots 
being  equal,  that  equation  is  \x—0(f}\n—0.  Hence  x— 0(p)=0 
is  an  equation  in  12.  Therefore  0(p)  is  a  number  in  12,  as  is 
also  its  equal,  /(«,  •••,  «„_!). 

Ex.  1.  In  Ex.  1,  §  151.  the  Galois  group  is  1,  («2«3),  (««i),  (««i)  • 
(«2«3).  The  roots  of  /(a;)  =  0  are  «,  «l5  «2,  «3-  Then  a2  +  4  a\  +  10  is 
a  function  in  tyu  of  two  roots  of  f(x)  =  0.  The  value  of  this  function  is 
50,  a  number  in  fyij ;  that  is,  belonging  to  the  domain  fi(i).  Performing 
the  substitutions  (««i),  we  get  «i2  -f  4  «  +  10,  which  still  equals  50.  The 
other  substitutions  do  not  affect  the  function.  This  illustrates  §  153. 

Ex.  2.  Using  the  group  and  roots  of  Ex.  1,  illustrate  §  153  by  the 
equation  («2  +  4  «i  —  24)2(«22  +  8  «3  —  60)3  =  0.  Here  the  left  member 
of  the  equation  is  owe  function,  and  the  number  in  n  is  0. 

Ex.  3.  /(x)  =  a;*  —  x2  —  2  =  0  has  the  Galois  domain  fyp),  where 
p  =  Vt  +  i,  pi  =  \/2  —  i,  p2  =  —  V2  +  t,  p3  =  —  V2  —  L  (1)  Express  each 
of  the  roots  of  f(x)  =  0  as  a  function  of  p.  (2)  Find  the  group  of  the 
domain.  (3)  Find  the  Galois  group  of /(x)  =  0. 

Ex.  4.  In  Ex.  3  show  that  /(«,  •••,  an-i)  =  «8  +  «i3  +  «28  +  «s8  ad- 
mits all  the  substitutions  of  the  Galois  group  ;  then  show  by  actual  sub- 
stitution that /(a,  •••,  «n_i)  is  a  number  in  fl(i).  This  illustrates  §  154. 

155.  Theorem.     A  group  G  is  a  Galois  group  of  the  equation 
f(x)  =  0  for  the  domain  Q  whenever 

(A)  Every  function  in  ft  of  the  roots  ai}  which  is  a  number  in 
12,  admits  the  substitutions  of  G,  and 

(B)  Every  function  in  O  of  the  roots  «„  which  admits  the  sub- 
stitutions of  G,  is  a  number  in  (2. 


166  THEORY    OF    EQUATIONS 

Firstly,  we  prove  that  every  substitution  of  G  belongs  to  the 
Galois  group. 

As  in  §  142,  select  appropriate  values  in  O  for  the  coefficients 
c,  cly  •  ••,  cn-1  so  that  distinct  values  for  p  are  obtained  for  every 
permutation  of  the  roots  a,  a1?  •  •  •,  «„_!  in  the  relation 


C«  +  Citti  -\  -----  h  <>„_!«„_!  =  P> 

Now  p  is  a  root  of  the  Galois  resolvent  g  (y)  =  0.  In  g  (p)  =  0 
substitute  for  p  its  value  in  I  and  we  get  a  function  in  O  of  a, 
ai>  •••>  ««-i>  which  equals  the  number  zero.  If  this  function 
satisfies  hypothesis  (A),  it  admits  any  substitution  s  of  the 
given  group  G.  But  by  I  this  substitution  changes  p  into 
some  distinct  value  pa.  Hence  g  (pa)  =  0,  and  pa  is  a  conjugate 
of  p.  But  the  substitution  (ppa),  which  corresponds  to  s,  is  a 
transposition  of  the  Galois  domain;  hence  s  belongs  to  the 
Galois  group,  and  G  is  either  the  Galois  group  or  one  of  its 
sub-groups. 

Secondly,  we  prove  that  the  Galois  group  is  G  itself. 
Suppose  G  embraces  j  substitutions,  namely, 

S)   "')  Sk)  Sj-D 

then  the  application  of  each  of  these  to  the  function  p  in 

I  yields  the  values  TTT 

P,  •••,  p{,  PJ-I. 

If  we  operate  with  any  substitution  sk  in  II  upon  any  value 
pt  in  III,  the  result  p'{  must  be  the  same  as  if  we  had  operated 
upon  p  directly  with  s(sk.  But  sfs*  must,  by  the  definition  of  a 
group,  be  one  of  the  substitutions  in  II  ;  hence  p'{  must  be  one 
of  the  values  in  III.  Thus  it  is  evident  that  the  operation 
with  sk  upon  every  value  of  III  causes  simply  a  permutation 
of  the  values  in  III.  Hence  a  function  g'(y),  defined  by  the 
relation  gl(y}  ^(y  _  p)(y  _  pj  ...  (y  _  ^ 

has  coefficients  of  y  that  are  each  invariant  under  the  substi- 
tutions of  G.  If  we  apply  to  each  of  these  coefficients  the 


NORMAL   DOMAINS  161 

hypothesis  (B),  each  of  them  is  a  number  in  (i.     Consequently 
g'(y)  is  a  function  of  y  in  O. 

Now  g'(y}  =  0  and  the  Galois  resolvent  g(y)  =  0  have  the  root 
p  in  common,  hence  (§  126)  the  degree  of  g'(y)  =  0  cannot  be 
less  than  that  of  g  (y)  =  0  ;  that  is,  j,  which  is  the  order  of  G, 
cannot  be  less  than  the  order  of  the  Galois  group.  Hence  the 
two  groups  are  the  same. 

156.  Theorem.  An  equation  is  reducible  or  irreducible  accord- 
ing as  its  Galois  group  is  intransitive  or  transitive. 

Let  /(a?)  =  F(x)  -h(x)  =  0, 

where  f(x)  =  0  is  reducible  and  /(»),  F(x),  h  (x)  are  functions 
in  O.     Let  the  roots  of  F(x)  =  0  be 


These  are  also  roots  of  f(x)  =  0,  which  has  the  following 
additional  roots  :  TT 

av)    "')   aj>   an-\- 

Now  it  is  evident  that  no  root  at  of  set  I  can  be  replaced  in  the 
equation  F(x)  =  0  by  a  root  a,  of  set  II,  for  «,.  is  not  a  root  of 
F(x)  =  0.  Yet  we  know  that  the  coefficients  of  x  of  F(x)  =  0 
admit  all  the  substitutions  of  the  Galois  group  of  f(x)  =  0 
(§  153).  Hence  this  group  can  have  no  substitution  which 
replaces  at  by  «,.,  and  the  group  is  intransitive  (§  102). 

Conversely,  if  the  group  P  is  intransitive  and  permutes  the 
roots  in  set  I  among  themselves  only,  so  that  at  will  not  be 
replaced  by  «,-,  then  the  product 

^(aj)  =  (a?  -  a)  (x  -  «j)  •  •  •  (x  —  av_^ 

admits  of  all  the  substitutions  of  P,  and  is,  therefore,  a  function 
of  x  in  O.  Hence  F(x)  is  a  factor  in  fi  of  /(a?),  and  /(a?)  =  0  is 
reducible. 

Ex.  1.  Illustrate  this  theorem  by  showing  that  the  Galois  groups  of 
Exs.  1  and  4  in  §  151  are  intransitive. 


168  THEORY   OP   EQUATIONS 

157.  Theorem.  An  imprimitive  domain  has  an  imprimitive 
group. 

Let  f(x)  —  0,  having  the  roots  a,  a1}  «»,  «n_1?  be  irreducible. 
Then  its  Galois  group  P  is  transitive  (§  156).  Let  the  do- 
main O(a)  be  imprimitive;  that  is,  let  it  possess  imprimitive 
numbers  which  are  not  all  numbers  in  O  (§  135).  If  N=  <f>(a) 
is  an  imprimitive  number,  then  its  conjugates  may  be  divided 
into  ni  sets  of  n2  equal  numbers  in  each  set,  so  that  n  =  nt  •  n2 
(§  138).  We  have  then  the  following  %  sets  of  roots  of  f(x)  =  0 
with  nz  roots  in  each : 

A=a    «      .••     are 


s  = 

so  that 


II 


are  numbers  conjugate  to  N. 

From  II  we  see  that  the  Galois  group  P  of  f(x)  =  0  must  be 
so  constituted  that  the  roots  of  each  set  A,  S,  •••,  S  are  inter- 
changed among  themselves  and  that  the  sets  A,  B,  •••,  S  are 
interchanged  bodily,  but  never  can  two  roots  of  the  same  set 
be  replaced  by  two  roots  belonging  to  different  sets.  Hence  P 
is  an  imprimitive  group  (§  103). 

Ex.  1.  Show  that  the  group  composed  of  the  powers  of  (0123)  is  an 
imprimitive  group. 

Ex.  2.  Show  that  any  cyclic  group  whose  order  is  not  prime  is  an 
imprimitive  group. 

158.  Theorem.  The  symmetric  group  of  the  nth  degree  is  the 
Galois  group  of  the  general  equation  f(x)  =  0  of  the  nth  degree  in 
the  domain  fi,  defined  by  the  coefficients  of  f(x). 


NORMAL   DOMAINS  169 

In  the  general  equation  f(x)  =  0  no  relation  is  assumed  to 
exist  between  the  roots;  that  is,  the  roots  are  taken  to  be 
independent  variables. 

In  all  cases  a  symmetric  function  in  O  of  the  roots  equals 
a  number  in  (i  (§  70).  If  it  be  granted  that,  for  the  general 
equation,  this  is  the  only  function  in  O  having  this  property, 
condition  A  of  §  155  demands  simply  that 

Every  symmetric  function  of  the  roots  shall  admit  the  sub- 
stitutions of  the  symmetric  group, 

and  condition  B  demands  that 

Every  such  symmetric  function  shall  equal  some  number  in  O. 

Both  statements  are  true.  Hence  the  symmetric  group  is 
the  Galois  group  of  the  general  equation. 

159.  Actual  Determination  of  the  Galois  Group.  In  Exs.  1  and 
4  of  §  151  we  determined  the  Galois  groups  of  easy  equations, 
for  the  domain  denned  by  the  coefficients  of  the  equation,  by 
the  aid  of  the  roots  of  the  equations.  When  the  roots  are  not 
known,  P  might  be  obtained  by  the  construction  of  the  Galois 
resolvent,  from  which  P  is  obtainable.  But  the  Galois  resolvent 
is  not  easily  constructed.  Practically  the  Galois  group  can  be 
ascertained  more  readily  from  the  theorem  about  to  be  deduced. 
It  is  well  to  remember  that,  when  f(x)  =  0  is  irreducible,  the 
degree  of  the  Galois  group  is  equal  to  the  degree  of  the  equa- 
tion. When  /(of)  =  0  is  reducible  and  the  factors  are  known, 
it  is  easiest  to  consider  the  equations  resulting  from  the  irre- 
ducible factors  of  /(#).  We  proceed  to  prove  the  following 
theorem,  in  which  M  is  any  function  in  fl  of  the  roots  a,  •••,  an_i, 
which  belongs  to  Q  as  a  sub-group  of  the  symmetric  group : 

If  a  function  M  is  a  number  in  O,  the  Galois  group  for  the 
domain  O  is  either  Q  or  one  of  its  sub-groups. 

Since,  by  hypothesis,  M  is  a  function  in  O  of  the  roots 
a,  «0  .-.,  «„_!,  which  is  a  number  in  O,  it  follows  by  §  153  that 


170  THEORY   OF   EQUATIONS 

M  admits  of  every  substitution  of  the  Galois  group.  By  defini- 
tion, M  belongs  to  Q ;  that  is,  there  are  no  substitutions  of  the 
roots,  except  the  substitutions  in  Q,  which  leave  M  unaltered 
in  value.  Hence  the  Galois  group  is  either  Q  or  one  of  its 
sub-groups. 

Ex.  1.    For  the  domain  fi(a. ...,  etn_1)  the  Galois  group  of  f(x)  =  0  is  1. 

Let  Q  =  1  and  M  =  c(t  +  •••  +  Cn-\(tn-\  be  a  function  in  O  of  the  roots, 
such  that  it  is  altered  in  value  for  every  interchange  of  the  roots.  Then 
M  belongs  to  Q,  and  is  a  number  in  the  given  domain.  Hence,  by  the 
above  theorem,  P  —  1  for  fi((li ...,  o^p. 

Ex.  2.    Find  the  Galois  group  of  the  cubic  «3  +  3  x2  -  6  x  +  1  =  0. 

The  discriminant  (§  35)  is  found  to  be  272.  By  §  77  the  alternat- 
ing function  of  a,  «i,  «2  equals  the  square  root  of  the  discriminant. 
This  function  admits  the  alternating  group.  See  Ex.  1,  §  100.  Take 
M  =  («- «i)(a- <x2)(«1  -  02)=  27,  Q  =  £3(3>,  and  Q  =  fl(1).  We  see 
that  M  is  unaltered  in  value  by  the  substitutions  of  GV3>,  but  that  its  alge- 
braic sign  is  altered  by  the  remaining  substitutions  of  GV3)-  Hence  M 
belongs  to  GV3>  ;  M  is  a  number  in  O(D.  Therefore  the  required  group  is 
either  G^3)  or  the  group  1.  By  §  54  we  see  that  the  equation  has  irra- 
tional roots  ;  hence  P  cannot  be  1,  it  must  be  GV3)  for  the  domain  fi(i). 

Ex.  3.   Find  the  Galois  group  of  Newton's  cubic 

X3  _  2  x  -  5  =  0. 

The  discriminant  is  not  a  perfect  square  ;  hence  P  =  G^l  for  Q(D. 
Ex.  4.    Show  that  P  =  GV3>  for  the  cubic 

x3  -  3(c2  +  c  +  l)x  +  (c2  +  c  +  1)(2  c  +  1)=  0 
and  the  domain  12(1,  c). 

Ex.  5.  Show  that  GV4)II  is  the  Galois  group  of  x*  +  1  =  0  for  the 
domain  fl(i). 

The  discriminant,  §  51,  is  256,  a  perfect  square.  Hence  the  alternating 
function  which  belongs  to  6ri2(4)  is  a  number  in  fl(i).  The  required  group 
is  either  G(i2(4)  or  one  of  its  sub-groups.  It  cannot  be  the  identical  group, 
because  the  roots  are  not  rational ;  it  cannot  be  G^W,  because  this  is 
intransitive,  while  x*  +  1  is  irreducible  (§  156).  Hence  the  group  is  either 
GI^>  or  #4<4>II.  We  see  that  y=  (a-  «i)  (03  — «s)  is  unaltered  by  GV4>II, 
but  is  altered  in  form  by  all  substitutions  not  in  G^WII.  The  resolvent 
cubic,  having  y  as  a  root,  is  y3  —  12  y  +  16  ^  0  (Ex.  17,  §  71).  Since  the 


NOEMAL  DOMAINS  171 

roots  of  this  resolvent  are  rational,  y  is  a  number  in  Oa).  Since  these 
roots  are  distinct,  y  is  altered  not  only  in  form,  but  also  in  value  by  sub- 
stitutions not  in  £4WII.  Hence  y  belongs  to  G^WII,  and  we  may  take 
y  =  M.  Hence  G^WII  is  the  required  group. 

Ex.  6.   Find  P  for  the  equation  (x2  +  2)  (x2  +  x  +  1)  =  0,  Oa). 

The  Galois  group  of  a;2  +  2  =  0  for  O(1)  is  P  =  1,  (««!).  The  equation 
x2  4-  x  +  1  =  0  gives,  for  0(1),  P>  =  1,  («2«3).  If  we  multiply  the  substi- 
tutions of  P  by  those  of  P1,  we  obtain  the  intransitive  group  1,  (rt«i),.  («2a3), 
(aai)(a2tt3)  =  (T4(4)III  as  the  required  group  for  the  domain  Oa).  See 
Ex.  6,  §  104. 

Ex.  7.  For  the  domain  0(1)  ,  x3  -  2  x  -  5  =  0  has  P  =  6V3).  Show  that 
for  the  domain  i  /7>  :?> 


Ex.  8.    For  the  given  domains  find  the  Galois  groups  of 

(a)  x2  +  5  x  +  6  =  0,  n(i). 

(6)  x2  +  5  x  +  5  =  0,  0(i). 

(c)  z*  +  H)  =  0,  0(ixio). 

(d)  (x  +  1)3  =  0,  0(1). 

(e)  x3  -  21  x  +  35  =  0,  O(D. 

(/)  x3  _  3(3  +  v/2)a;  +  7  (1  +  V5)  =  0,  0(1,  rf). 

(g)  x*  +  x3  +  x2  +  x  +  1  =  0,  Oa). 

(A)  (x2  +  5)  (x3  -  21  x  +  35)  =  0,  0(1),  also  O(1,  vt).    See  Ex.  7,  §  104. 

(0  x6-!  =(x  +  l)(x-l)(x2  +  x  +  l)(x2-x  +  l)  =  0,  0(1). 

(*)  x'2  -  1  =  0,  0(1). 

(0  x1  +  (a  +  6)x2  +  ab  =  0,  0(1,  a,  6). 

(m)  x3  -  2  =  0,  0(1). 

(n)  x*  +  4  x3  +  6  x2  +  4  x  +  2  =  0  for  Oa). 

Ex.  9.    Find  a  general  expression  for  the  equation  of  the  fourth  degree 

whose  Galois  group  is  G&(4).     Assume 


(a  -  «2)2  +  («!  -  «3)2  =  8  c, 
[(a  -  «2)2  -  («i  -  «3)2]2  =  64  6, 
[(a  -  aa)2  -(«i  -  «3)2][«  -  «i  +  «2  -  «3]= 
where  6,  c,  d  are  rational  numbers  and  b  is  not  a  perfect  square.     These 


172  THEORY    OF    EQUATIONS 

assumptions  are  justified  by  the  fact  that  the  left  member  of  each  equa» 
tion  is  a  function  which  belongs  to  G^*\  §  154-     We  get 

(a  -  «2)2  =  4(c  +  V&),  («i  -  «3)2  =  4(c  -  V&), 
a  —  «i  +  «2  —  «s  =  4  dVfc, 

«  +  #1  +  «2  +  «3  =  4  •&!. 


Hence  a  =  61  -f  tf  V6  +  Vc  +  V6,  «2  =  61  +  d  Vb  -  Vc  +  V&, 

-  Vc  -  V6. 

Diminishing  each  root  by  61  and  forming  the  quartic,  we  obtain  the 
y*  -  2(bcP  +  c)?/2  -  4  bdy  +  (bcP  -  c)«  -  6  =  0. 


Ex.  10.  The  quartic  whose  Galois  group  is  G^WIII  is  the  reducible 
equation, 

x*  -  2(c2  -f  d)xz  -  4  cex  +  (c2  -  <Z  +  e)(c2  -  d  -  e)  =  0, 

where  (<?  +  e)  and  ((7  —  e)  are  not  perfect  squares. 
Derive  this  by  assuming 

«i  +  «2  —  «s  —  «4  =  4  c, 
(«!  -  «2)2  +  (OB  -  «4)2  =  8  d, 
(«i  -  «2)2  -  («3  -.«4)2  =  8  e. 

Ex.  11.   Find  a  general  expression  for  equations  of  the  fourth  degree 
having  the  Galois  group  G£*>1.     Use  the  functions 


+  i&z  -  a3-  z«4) 

-  «2  +  «3  -  «4)2, 

—  ia%  —  03 

-  «2  +  «3  - 


and  impose  upon  the  letters  which  appear  in  the  expressions  for  the  co- 
efficients of  the  quartic  no  other  conditions  than  that  they  shall  be  rational 
and  one  of  them  shall  not  be  a  perfect  fourth  power.  See  Ex.  3,  §  176. 

Ex.  12.  Show  that,  if  the  roots  of  the  cubic  in  Ex.  11,  §  71,  are  all 
rational,  the  Galois  group  of  the  quartic  having  the  roots  o,  /3,  7,  8  is 
either  G^Il  or  one  of  its  sub-groups. 

Consider 


NORMAL   DOMAINS  173 

Ex.  13.  The  product 

(«1  +  «2  -  «3  -  «4)(«1  -  «2  +  «3  -  04)(«1  -  «2  -  «3  +  «4> 

is  a  symmetric  function  of  cti,  «2,  «s,  «4.     The  square  of  the  product  ot 


the  first  two  factors  belongs  to  G^WII.  To  find  the  general  quartic  having 
Gt^Il  as  its  Galois  group,  we  may  therefore  assume  the  factors  to  equal, 
respectively,  Vft,  Vc,  dVbc,  where  6,  c,  d  are  rational,  but  where  be  is 
not  a  perfect  square. 

The  required  equation,  deprived  of  its  second  term,  is 


y*  -  2(6  +  c  +  bed2)!/2  -  8  bcdy  +  (b-c-  fccd2)2  -  4  bcW  =  0. 

Ex.  14.  Show  that  x4  -f  2  6x2  +  c  =  0  has  the  group  G8(4)  when  b  and  c 
are  subject  only  to  the  condition  that  6'2  —  c  is  not  the  square  of  a  number 
in  0(i,  &,  c)- 

Ex.  15.  Show  that  x*  +  2  6x2  +  c  =  0  has  the  group  G^4'!!  when  c,  but 
not  62  —  c,  is  the  square  of  a  number  in  fyi,  5). 

Ex.  16.  Show  that  s*  -  8  Sx2  +  8  S*  -  8  8*  =  2,  where  £  is  any  number 
in  0(i),  has  the  group  G^WI.  See  Ex.  11. 


CHAPTER  XV 

REDUCTION  OF  THE  GALOIS  RESOLVENT  BY  ADJUNCTION 

160.  Definition   of   M.     Let    the    Galois    group   P   (of    the 
order  p~)  of  the  equation  f(x)  —  0,  having  the  roots  a,  a1}  •••,  an_lf 
possess  a  sub-group  Q  of  the  order  q,  where  p  =  qj,  j  being  the 
index  of   Q  under  P.     For  the  purposes  of  the  theorems  in 
succeeding  chapters,  we  define  M  nearly  as  in  §  159. 

Let  Mbe  any  function  in  Q  of  the  roots  a,  •••,  «a_j  which  belongs 
to  Q  as  a  sub-group  of  P  (§  111). 

161.  Theorem.     By  operating  upon  M  with  the  substitutions  of 
P  we  obtain  j  distinct  values  of  M  which  are  roots  of  an  irreducible 
equation  of  the  jth  degree  in  O. 

If  Ms  a  substitution  of  the  Galois  group  P  which  does  not 
occur  in  the  sub-group  Q,  and  if  s,  sv  ••-,  s^  be  the  substitutions 
of  Q,  then  by  the  definition  of  a  group, 

st}  sj,  •  ••,  sg_i£, 

are  all  substitutions  of  P.  But  the  substitutions  srt  in  I,  when 
applied  to  M,  all  produce  the  same  effect,  for  in  any  case  we  may 
operate  with  the  product  sjt,  by  first  operating  with  sr  and  then 
upon  the  result  with  t.  By  hypothesis,  operating  with  sr  upon 
M  produces  no  change  whatever,  hence  srt  produces  always  only 
the  result  due  to  t  alone. 

By  hypothesis  it  follows  that,  as  t  does  not  occur  in  the  sub- 
group Q,  t  operated  upon  M  gives  us  a  new  value  M^ 

From  §  106  we  see  that  there  are  as  many  sets  of  substitu- 
tions I  in  the  group  P  as  q  is  contained  in  p ;  namely,  j  sets. 
The  substitutions  of  any  one  set  applied  to  M  all  give  the  same 
value  for  M,  but  no  two  sets  yield  the  same  value. 

174 


GALOIS    RESOLVENT   BY    ADJUNCTION  175 

For  suppose  sjt  and  srtk  yielded  the  same  value  for  M ;  that 
is,  suppose 

Mt  =  M  operated  upon  by  srt{ 

and  M(  =  M  operated  upon  by  sj^ 

then,  operating  with  (V.-T1  upon  Mi  would  give  M  =  M  operated 
upon  by  (V*)^)"1- 

That  is,  (sr/t)(sr^)~1  is  a  substitution  contained  in  the  group  Q 
and  is  equal  to,  say  sm.  If  sm  =  (5,4)  («,.«,.)  -1,  then,  operating  with 

sr^,  we  get 

srtk  =  S.M  =  sm't{, 

where  sm'  is  a  substitution  in  Q.  Since  the  effects  of  srtk  and 
sm'ti  upon  IT  are  the  effects  of  tt  and  £,  alone,  it  follows  that 
tk  =  t»  which  is  contrary  to  supposition.  Hence  srtf  and  srtk  must 
yield  different  values  when  applied  to  M. 

The  function  <£(?/)  =  (y  —  M}(y-M^--(y-  J/,_j)  is  now 
seen  to  be  invariant  under  any  substitution  of  P. 

The  coefficients  of  y  in  <f>(y),  obtained  by  performing  the  indi- 
cated multiplications,  are  symmetric  functions  of  M,  M^  •••, 
MJ_I  and,  therefore,  by  the  definition  of  M,  functions  in  O  of 
the  roots  of  f(x)  =  0,  functions  which  admit  of  the  substitu- 
tions of  the  Galois  group  P.  Hence  these  coefficients  are 
numbers  in  fi  (§  154). 

To  prove  the  irreducibility  of  <£(?/),  assume  that  0(y)  is  any 
function  of  y  in  fl,  which  vanishes  for  y  =  M.  Then  0(M )=  0. 
Since  6(M)  must  admit  all  the  substitutions  of  the  Galois 
group  (§  153),  we  must  have  0(Mi')=  0,  where  i  has  any  value 
0, 1,  2,  ••-,  (j  —  1).  Hence  6(y)  cannot  be  of  lower  degree  than 
the  jih.  As  all  the  roots  M,  Ml}  •••,  Mj^  of  </>(#)  =  0  satisfy 
%)  =  0,  0(y)  is  divisible  by  <£(»• 

Now,  if  <£(?/)  were  reducible,  one  of  its  factors  would  vanish 
for  y  =  M.  Since  6(y)  may  be  any  algebraic  function  in  13 
which  vanishes  for  y  =  M,  let  0(y)  represent  this  factor.  Then 
it  would  follow  that  this  factor  would  be  divisible  by  the  whole 
product  <f>(y),  which  is  impossible.  Hence  <f>(y)  is  irreducible 


176  THEORY   OF   EQUATIONS 

162.  Theorem  of  Lagrange  as  generalized  by  Galois.  Any 
number  in  the  Galois  domain  which  admits  the  substitutions  of 
the  group  Q  is  contained  in  the  domain  fi(JO. 

In  §  161  we  saw  that  M,  a  function  which  belongs  to  Q, 
assumed  the  following  distinct  values,  when  operated  on  by  the 

substitutions  of  P  '•         M  nr          n/r  T 

ja}  MX  '•',  MJ-I. 

Let  M1  be  any  function  in  O  of  the  roots  «,  •••,  «„_!  which 
admits  the  substitutions  of  Q.  Let  any  substitution  of  P 
which  changes  M  into  Ma  change  M1  into  M' „  then  we  get  the 
following  values,  corresponding  to  those  in  I, 

M',  M\,  ..-,  M'^.  II 

These  are  not  necessarily  distinct. 

Accordingly  when  upon  the  series  of  numbers  I  and  II  we 
operate  with  a  substitution  of  P,  there  occurs  a  permutation  in 
each  series,  but  such  that  if  Mt  changes  to  Mr)  then  M' ,  changes 
to  M ',.. 

Denning  <£(?/)  as  in  §  161,  consider  the  function 


which  is  an  integral  function  of  y  of  the  (j  —  l)th  degree. 
This  function  is  invariant  under  all  substitutions  of  P.  Hence 
it  is  a  function  in  O.  Take  y  =  M.  Remembering  that  <j>(y) 
has  no  equal  roots,  we  have  (reasoning  as  in  §  142) 


where  <j>'  indicates  the  first  differential  coefficient  of  <£  with 
respect  to  y.     Thus  M'  is  a  number  in  the  domain  O(Jf). 

Ex.  1.  Find  the  value  of  a  root  a  of  the  equation  x2  +  2  =  0  in  terms 
of  a  —  «i,  it  being  given  that  P  =  1,  (««i). 

If  we  take  Q  =  1,  we  see  that  M=ct.  —  a\  is  a  function  which  belongs 
to  Q  and  that  M'  =  a  is  a  function  which  admits  Q.  We  find  M\  =  «i  —  «, 


GALOIS   RESOLVENT   BY    ADJUNCTION  177 


-2aa1=-8,    0'(y)  =  2  «/.      Hence  a  =  4>(M)/<f>'(M)  =-  4/M.     The 
correctness  of  this  result  is  easily  shown. 

Ex.  2.  For  the  equation  x'2  +  ax  +  b  =  0,  having  the  group  P  =  1, 
(««i),  find  a3  —  «i2  as  a  function  of  «  in  Ii(1). 

Take  §  =  1,  M=  a,  M'  =  a3  -  ai2,  then  4>(z/)  =  (3  aft  +  2  6  -  a?  -  a3)y 
+  3  ab  +  2  b2  -  a*b  -  a3,  0'(«/)  =  2  y  +  a.  Hence 

Jf'  =  [(3  a&  +  2  5  -  a2  -  a3)  M  +  3ab  +  2b2-  a'2b  -  a3]  -  (2  Jf  +  a). 

Ex.  3.  Find  the  value  of  [w,  a]3  for  the  cubic  x3  +  a\xz  +  azx  +  as  =  0 
in  terms  of  the  alternating  function  («  —  «!)  (a  —  «2)  («i  —  «;>)=  VZ>. 
Let  ilf  =  \/Z).  then  i  M=  -  VD. 

We  have  M'  =  [w,  a]3,  Jf'  1  =  [a>2,  a]3,  <t>(y)  =  y*  -  Z>, 

*(y)  =  ?/(,¥'  +  Jf  0  f  V5(  Jf  -  Jf  0.     By  §  71,  Ex.  15, 
M'  +  M'i  =  -  2  «!3  -V  9  aia-2  -  27  ag.    We  find  M  -  M^  =-Si  Vz~D, 

(-  2  ai3  +  9  ai«2  -  27  a3  -  3  i  V3Z>), 
=2  \/^,  Jf'  =  i(-  2  ai3  +  9  aia2  -  27  as  -  3  f  V3l>). 
See  also  the  solution  in  §  173. 

Ex.  4.  For  the  quartic  te4  +  4  ft^3  +  6  62a;2  +  4  63x  +  &4  =  0,  find  the 
value  of  M'  =  («  +  «2)  («!  +  «3)  in  terms  of  M,  where 

16  MI  =  (a  -  a  +  «2  -  «a)2. 

Both  Jlf  and  M'  belong  to  the  group  Gs(4)-  Notice  that  M  is  a  root  of 
the  cubic  III,  §  62.  See  also  §  109.  Hence  that  cubic  is  00)  =0.  We  find 

162<i>(2/)  =  162  (Mf  +  M'i  +  M'  n)y2  -  16({.l/!  +  Mn}M> 
+  {M+  JfiJJf'n)  y  +  M^M^l'  +  JOfnJf'i  +  J/Mi 

=  162  •  2  S«1a2  •  y2  -  16  (4  £««!  .  S«2  - 
+  (2  2a5«!  -  6  2a*ai02  +  4  S«3«i2a2 

In  Ex.  16,  §  71,  the  values  of  the  symmetric  functions  occurring  here  are 
given. 

Ex.  5.    Complete  the  computation  in  Ex.  4  for  the  special  quartic 
=0.     We  obtain  *(y)  =  12  j/2  -  16  y  -  3, 


178  THEORY    OF    EQUATIONS 

163.  Reduction  of  Galois  Group.  If  we  adjoin  to  Q  a  function 
M,  the  Galois  group  reduces  to  Q. 

Firstly,  each  function  in  !2(JO  of  the  roots  a,  aly  •••,  an_l  of 
the  original  equation  f(x)  =  0,  which  equals  a  number  in  O(Jf), 
admits  the  substitutions  of  Q;  for,  this  number  in  fi(ar)  is  a 
function  of  M,  and  M  admits  all  the  substitutions  of  Q. 

Secondly,  each  function  in  &(M)  of  the  roots  a,  •••,  an_l}  which 
admits  the  substitutions  of  Q  is  by  §  162  a  number  in  O(JO. 

But  these  are  the  two  characteristic  properties  of  the  Galois 
group  in  the  domain  fi(JO  (§  155).  Hence  Q  is  the  Galois  group 
of /(#)  =  0  in  the  new  domain  Q(JO. 

This  reduction  of  the  order  of  the  Galois  group  from  p  to  q 
(§  160)  was  effected  by  the  adjunction  of  J/,  the  root  of  an 
auxiliary  equation  of  degree  J  (§  161). 

*  Ex.  1.  Given  that  x*  -f  Xs  +  1  =  0  has  the  Galois  group  G2^  for  fid). 
Adjoin  in  succession  four  irrationals  M  and  show  that  the  Galois  group  is 
reduced  and  the  domain  is  enlarged  as  indicated  below. 

M  P          <£(?/)  =  0,  §  161  Domain 

<?24(4>  %) 

VD  Gi<t<V     Z>=229 

y=(a-ai)(02-as)  W>II  ys-12y+V229=0,  §  71,  Ex.  17 


Show  that  y  involves  the  irrational  v  12V—  3  —  4 \/229. 

Ex.  2.     Show  that  the  roots  of  the  quartic  in  Ex.  1  can  be  expressed 
rationally  in  terms  of  the  roots  of  the  quadratics  in  z  and  w. 

*  Ex.  3.   Apply  the  process  of  Ex.  1  to  the  quartic 
xl  +  ai«8  +  a2«2  +  asx  +  04  =  0 

and  deduce  the  successive  resolvent  equations  <£(?/)  =  0 ;  viz., 
D  =  256(73-27t/2)   (§  51),  y3  -  12  7  +  \/Z>  =  0, 

=  72  ciV-  192  a2t7+  144  yj+  8  7y2  +  y  VD  -  64  /*, 

W2  —  210  +  W  =  0. 


GALOIS    EESOLVENT    BY    ADJUNCTION  179 

164.  A  Resolution  of  the  Galois  Resolvent.  Let  the  Galois 
resolvent  g(y)  =  0  have  a  root  p.  If  we  effect  upon  p  the  sub- 
stitutions st  of  the  sub-group  Q,  one  at  a  time,  we  get  the  values 

where  Pi  is  gotten  by  operating  upon  p  with  the  substitution  st. 
If  upon  the  p's  in  I  we  effect  any  substitution  of  the  group  Q, 
the  pt  in  I  simply  undergo  a  permutation ;  for,  each  result  thus 
obtained,  being  derived  from  p  by  effecting  two  substitutions 
in  succession,  is  equivalent  to  p,  operated  upon  by  that  substi- 
tution of  Q  which  is  the  product  of  those  two  substitutions. 
Hence,  ^  M^  =_  ^  —  p)(y  —  p^...(y  —  Pq_^y  H 

is  invariant  under  Q,  and  the  coefficients  of  y  in  expression 
II  are  numbers  in  O(Jf),  §  162.  By  the  notation  g(y,  M)  we 
mean  here  a  function  of  y  in  which  the  coefficients  of  y  are 
numbers  in  fi(ar). 

Now  g(y,  M)  is  a  divisor  of  g(y)  in  the  domain  Q(JO,  for  the 
former  is  of  degree  q,  the  latter  of  p,  and  p  —jq,  §  160. 

If  upon  II  we  effect  a  substitiition  t  which  occurs  in  P,  but 
not  in  Q,  we  get 


The  values  p(t),  p^,  •••,  /o?_i(O  are  roots  of  g(y)  =  0,  hence  III  is 
also  a  divisor  of  g(y). 

Two  sets  of  roots  p(t\  •«•,  pq~l(t)  obtained  from  two  distinct  sub- 
stitutions t,  are  either  indentical  or  they  have  no  root  in  common. 
Consequently,  two  distinct  functions  g(y,  M,~)  have  no  common 
factor,  and  we  have  the  resolution  into  distinct  factors 

9(y)  =  9(y,  W  '  9(9,  J*i)  ».  g(y,  3fy_,).  IV 

It  is  to  be  noticed  that  in  this  resolution  the  factors  g(y,  M^)  do 
not  usually  belong  to  the  same  domain  ;  they  belong  respectively 
to  the  domains  O,(M^  &(*»  •••,  fyjr  ^  Another  resolution  of 
g(y)  is  possible,  in  which  all  the  factors  belong  to  the  same 
domain  O. 


180  THEORY    OF    EQUATIONS 

165.  Adjunction  of  Any  Irrationality.  If  by  the  adjunction 
of  any  irrational  X  to  O  ice,  obtain  a  domain  Q,^,  in  ichich  the 
Galois  resolvent  g(y~)  =  0  becomes  a  reducible  equation,  so  that 

9i(y,  ^)  =  (y  -  p)(v  -  p.)  -  (y  -  p}-.) 

is  an  irreducible  factor  of  g(if)  in  Ou-,  of  the  degree  q,  then  in  this 
new  domain  the  Galois  group  is  reduced  to  the  sub-group 


Adjoin  X.     Since  g(y)  =  0  is  a  normal  equation  in  fi,  §  146, 
we  have  p4  =  <j>t(j>)-     In 

9i(y,  x)  =  (y-  p)(y  -  pi)  •••  (y  -  p,-0  =  o  i 

write  <£,•(?/)  in  place  of  y  ;  we  obtain  a  new  equation  in  y,  viz., 

-  p,.^)  =  0.      II 


As  I  is  irreducible  in  O  and  I  and  II  have  a  root  p  in  com- 
mon, all  the  roots  of  I  satisfy  II.  Let  ph  be  any  root  of  I  ;  then 
putting  y  =  ph,  one  of  the  factors  in  II  must  vanish  ;  say,  the 
factor  <k(pA)  —  Pk. 

We  have  now  the  relations 

Pi  = 


Hence  the  equality  of  the  substitutions 

W*)  =  (/>/>*)• 
Multiplying  by  (pp^,  we  have 


or  pPk) 

That  is,  the  product  of  any  two  substitutions  in  the  set 
1,  (ppi),  •••,  (ppg-i)  is  equal  to  one  of  the  substitutions  in  the  set. 
Hence  they  form  a  group,  §  95.  Call  this  sub-group  Q. 


GALOIS    RESOLVENT   BY   ADJUNCTION  181 

Equation  I  is  the  Galois  resolvent  of  f(x)  =  0  for  the  domain 
fi(jr)  ;  for  this  equation  is  by  hypothesis  irreducible  in  D(X),  and 
the  two  other  conditions  are  satisfied,  because  of  the  relation 

Q(a,.»,a,,_1)   =  O(p)   =  O^.),    §    145. 

Hence  Q  is  the  Galois  group  of  f(x)  =  0  in  the  domain  O(x). 

166.  M  a  Function  of  X.  M  can  be  expressed  as  a  function  in 
O  of  any  irrational  X  which  reduces  the  Galois  group  to  Q. 

We  have  seen  that  g^y,  X)  is  a  function  in  £l(x)  of  y,  whose 
coefficients  admit  the  substitutions  of  the  sub-group  Q.  Since 
M  belongs  to  Q  and  these  coefficients  admit  Q,  the  coefficients 
are  numbers  in  Q(JO,  §  162.  Hence  we  may  express  the  product 


as  a  function  of  y  and  X  and  designate  it,  as  above,  by  g^y,  X), 
or  we  may  express  it  as  a  function  of  y  and  M  and  designate 
it  by  g(y,  M).  We  have  then 


Now  M  is  the  root  of  an  irreducible  equation  in  O  of  degree  j, 
§  161  ;  namely,  the  equation 


of  which  the  other  roots  are'  Ml}  M2,  •••,  M^.     By  §  164  we 

have 

g(y)  =  g(y,  M)  .  g(y,  M$  -  g(y,  M^  III 


The  equation  I  is  not  satisfied  when  in  the  left  member  we 
substitute  for  M  one  of  its  other  conjugates  ;  for,  supposing  it 
were,  it  would  follow  that  g(y,  M)  is  equal  to  one  of  the  other 
factors  in  the  right  member  of  III,  a  conclusion  at  variance 
with  the  fact  that  g(y),  being  irreducible  in  Q,  can  have  no 
equal  roots. 


182  THEOKY    OF    EQUATIONS 

It  is,  therefore,  possible  to  assign  to  y  such  a  rational  value 

that  the  equation 

9(y,*)-9i(y,X)=Q,  IV 

in  which  z  is  regarded  as  the  unknown  quantity,  has  only  one 
root  in  common  with  equation  II  ;  namely,  z  =  M. 

The  H.  C.  F.  of  II  and  IV  is  consequently  a  binomial,  linear 
with  respect  to  z.  Since  the  coefficients-  of  z  in  both  II  and  IV 
are  numbers  in  O(X),  and  the  process  of  finding  the  H.  C.  F. 
includes  only  operations  of  subtraction,  multiplication,  and  divi- 
sion, and  thereby  never  introduces  new  irrationals,  it  follows 
that  the  H.  C.  F.,  z  —  M,  is  a  function  in  O(x).  In  other  words, 
J\fis  a  number  in  O(J),  and  therefore  a  function  in  O  of  X. 

COROLLARY  I.  The  domain  Q(JO  of  degree  j  is  a  divisor  of  the 
domain  O(x),  since  every  mimber  in  O(Jf)  is  a  function  in  fi  of  X. 

COROLLARY  II.  The  number  X  is  a  root  of  the  irreducible 
equation  h(y~)  —  0  of  the  same  degree  as  that  of  the  domain  f)(x), 
§  138.  Hence  the  degree  ofh(y~)  =  0  is  a  multiple  ofj,  the  degree 
of  equation  II. 

COROLLARY  III.  If  X  is  taken  as  a  function  in  O  of  M,  then 
O(X)  and  O(Jf)  are  identical. 

COROLLARY  IV.  The  reduction  of  the  Galois  group,  caused 
by  any  irrational  X  which  is  not  a  number  in  the  Galois  domain, 
can  be  effected  equally  well  by  some  number  M  which  is  in  the 
Galois  domain.  That  is,  every  possible  reduction  of  the  Galois 
group  may  be  effected  by  the  adjunction  of  some  number  belonging 
to  the  Galois  domain. 

The  numbers  in  the  Galois  domain  of  the  equation  /(#)  =  0 
are  called  by  Kronecker  the  "natural  irrationalities"  of  /(a;)=0. 
The  corollary  may  now  be  stated  thus  :  Every  possible  reduction 
of  the  Galois  group  may  be  effected  by  the  adjunction  of  a  natural 
irrationality. 


Ex.  1.    In  Ex.  1,  §  163,  adjoin  to  0(1),  X=  "VV^.     Here  X  admits  the 
substitutions  of  the  alternating  group,  and  the  Galois  group  is  reduced 


GALOIS   RESOLVENT   BY   ADJUNCTION  183 

to    (rj2(4)-      Now  X  does   not  occur  in  the  Galois  domain  fi(a(  ( 

=  0(i.  \/»,  »,  z,  »)  and  's>  therefore,  not  a  natural  irrationality.    The  reduction 
brought  about  by  X  can  be  effected  by  VZ>,  which  is  a  number  in  the 
Galois  domain,  hence  is  a  natural  irrationality.     This  illustrates  Corollary 
IV. 
The  relation  VZ>  =  _3TB  illustrates  the  theorem  itself.    We  have 

+  VI>)  =  0,  or  y*  =  D. 


.  2/2  =  -\/-VD>  and  we  S 
=  0,  or  yin  —  D.    This  illustrates  Corollaries  II  and  I. 

Ex.  2.    If  the  group  P  of  an  equation  is  Gs(4\  illustrate  the  above 


theorem  and  corollaries  by  taking  X=     (aai  —  aaas^Ctteca+ttiOs)2.    See 
Ex.  6,  §  113. 


CHAPTER   XVI 

THE  SOLUTION   OF  EQUATIONS  VIEWED  FROM   THE  STAND- 
POINT  OF   THE   GALOIS   THEORY 

167.  General  Plan.  Quadratic  Equation.  The  problem,  to 
solve  an  algebraic  equation,  is  replaced  in  the  Galois  theory 
by  another  problem,  to  bring  about  a  reduction  of  the  Galois 
group  and  a  lowering  of  the  degree  of  the  Galois  resolvent  by 
the  successive  adjunction  of  simple  algebraic  numbers.  If  a 
function  M  is  adjoined  to  O,  the  Galois  group  is  reduced  to  Q. 
It  becomes  necessary  to  determine  the  numerical  value  of  M 
for  the  given  equation  f(x~)  =  0.  This  we  endeavor  to  do  by 
the  construction  and  solution  of  an  auxiliary  equation  of  the 
degree  j,  where  j  is  the  index  of  Q  under  P.  The  roots  of 
this  auxiliary  equation,  or  resolvent,  are  the  required  values  of 
the  conjugates  of  M.  This  same  process  is  repeated  upon  the 
reduced  Galois  group  until  this  group  finally  becomes  1.  Then 
the  enlarged  domain  contains  the  roots  of  the  given  equation, 
and  the  values  of  the  roots  may  be  found  in  terms  of  the 
numbers  M,  M',  •••  which  have  been  adjoined  to  the  original 
domain. 

Quadratic  Equation.  The  Galois  group  of  x2  -f  cttf  +  «2  =  0 
is  the  symmetric  group  Gr2(2),  §  158.  Its  only  sub-group  is  1, 
§  104,  whose  index  j  =  2.  Take  M=a  —  al  Its  other  con- 
jugate value  is  Ml  =  «j  —  a.  M  and  Ml  are  roots  of  the  equation 
2=a12-4a2,  §  161.  We  get  y=±Vaf^4a, 


as  the  values  of  M  and  M^.  After  adjoining  M,  the  Galois 
group  is  1  ;  the  enlarged  domain  is  O(1(  „  ,  „  ,  y/a2_4a  )t  We  know 
that  a  +  «!  =  —  a1  and  a  —  «j  =  Vc^2  —  4  a%.  Hence 


2  a  =  —  «!  +  V«i2  —  4  az  and  2  «j  =  —  at  —  V«i2  —  4  a-j 
184 


THE   SOLUTION    OF   EQUATIONS  185 

Theoretically  there  is  an  infinite  number  of  ways  of  solving 
the  quadratic,  because  there  is  an  infinite  number  of  functions 
M  to  choose  from.  Thus  we  may  take  M  =  S  («  —  «1)2n+1, 
where  n  may  be  any  value  which  gives  M  and  Ml  distinct 
values,  and  S  is  any  symmetric  function  of  a,  «x. 

168.  Cubic  Equation.  From  the  point  of  view  of  the  Galois 
theory  the  solution  given  in  §  59  may  be  outlined  as  follows  : 
The  change  from  x  to  z  is  an  operation  which  does  not  alter  the 
domain.  The  same  is  true  of  the  change  from  z  to  x,  after  z 
has  been  found  ;  also  of  the  substitution  of  u  +  v  for  z,  and  its 
inverse,  and  of  the  elimination  of  v.  The  solution  of  the  cubic 


may  be  exhibited  thus  (where  VA  =  V— 

=  o,  §161  M  p 


The  numbers  adjoined  to  O'  are  determined  by  the  roots  of 
two  resolvent  equations  <£  (y)  —  0,  the  first  a  quadratic,  the 
second  a  pure  cubic  equation. 

169.   Quartic  Equation.     We  give  here  those  steps  in  the  solu- 

tion given  in  §  62  which  involve  an  extension  of  the  domain. 

Welet     l6u=(a-al  +  a2-a3Y,  16  v  = 

16  w  =  (a  —  «i  —  «2  +  «s)2- 

<KJ,)=O  M 


«'(«, 
=  l,  (Cd) 


186  THEOEY    OF   EQUATIONS 

Since  Cr4(4)III  is  an  intransitive  group,  the  quartic  can  be 
factored  in  the  domain  tl\u^-v).  The  two  quadratic  equations 
thereby  obtained  have  as  Galois  groups  1,  (ab),  and  1,  (cd), 
respectively.  From  VI,  §  62,  we  see  that  n'(S/-iV/v)  =  n'(V/-(N/v). 
Hence  it  is  not  necessary  to  adjoin  more  than  one  of  the  two 
irrationals  Vw,  V*o. 

The  quartic  offers  a  better  exhibit  of  the  Galois  theory  than 
did  the  quadratic  and  cubic  equations,  because  not  only  may 
we  select  a  great  variety  of  different  functions  M  at  each 
adjunction,  but  we  may  select  different  groups.  In  the  above 
solution  the  series  of  groups  taken  is  6r24(4),  G&w,  (74(4)III, 
G  =.  (1,  («&)),  Cr  =  1,  but  another  series  may  be  chosen,  viz. 
G»w,  GW4)>  #4(4)H,  G?\  I-  In  Exs.  1  and  3,  §  163,  a  solution 
of  the  quartic  is  outlined,  in  which  this  series  of  groups  is  used. 

Again,  we  may  effect  a  solution  by  first  adjoining  a  function 
that  belongs  to  the  cyclic  group  6?4(4)I  ;  say, 


y  =  aa 

To  be  sure,  the  first  resolvent  equation  <f>  (y)  =  0  will  be  of 
the  sixth  degree,  but  it  can  be  treated  as  an  equation  of  the 
third  degree  and  a  quadratic. 

The  number  of  different  solutions  of  cubic  and  quartic  equa- 
tions which  have  been  given  since  the  time  of  Tartaglia  and  Car- 
dan is  enormous.  For  information  on  different  solutions  consult 
L.  Matthiessen,  Grundzilge  der  Antiken  u.  Modernen  Algebra. 

It  would  seem  that  the  above  mode  of  procedure  should  lead 
to  solutions  of  the  general  quintic  equation.  But  an  unexpected 
difficulty  arises  in  our  inability  to  solve  all  the  resolvent  equa- 
tions. There  arise  resolvents  of  higher  than  the  fourth  degree. 
The  Galois  theory  will  furnish  proof  that  the  solution  by  radi- 
cals of  the  general  quintic  and  of  general  equations  of  higher 
degrees  is  not  possible.  In  the  remaining  chapters  we  shall 
demonstrate  this  impossibility  and  discuss  the  theory  of  special 
types  of  equations  of  higher  degree  which  can  be  solved 
algebraically. 


CHAPTER  XVII 

CYCLIC  EQUATIONS 

170.  Definition.  A  cyclic  equation  is  one  whose  Galois  group 
is  the  cyclic  group,  §  101.  Kronecker  called  such  equations 
"einfache  Abel'sche  Gleichungen." 

A  quadratic  equation  is  cyclic  ;  for  the  Galois  group  is  the 
symmetric  group  6r2(2),  which  is  at  the  same  time  the  cyclic 
group  of  the  second  degree. 

The  general  cubic  is  not  a  cyclic  equation  in  the  domain  defined 
by  its  coefficients  ;  for  its  Galois  group  is  6r6(3),  which  is  not  a 
cyclic  group.  However,  if  we  adjoin 

VZ)  =  (a  —  «i)(a  —  02)  («i  —  «<}), 

the  Galois  group  becomes  (§  163)  G3(S),  which  is  cyclic.     Hence 
the  general  cubic  is  cyclic  in  the  domain  n(<v  „  aji  ^ny 

TJie  general  quartic  is  not  a  cyclic  equation  in  the  domain 
defined  by  its  coefficients,  but  if  we  adjoin  a  function  which 
belongs  to  the  cyclic  group  G4WI,  the  equation  is  cyclic  in  the 
new  domain.  One  such  function  that  may  be  adjoined  is 

M  =  aa*  +  «!«./  +  «2«32  +  «3«2. 
If  n  is  a  prime  number, 


is  a  cyclic  equation  in  the  domain  O(1).     For,  §  130,  this  equa- 
tion is  irreducible.     The  cyclic  function 


is  seen  by  the  relations  w2  =  w,2,  w3  =  cV,  etc.,  to  be  equal  to 
the  sum  of  the  roots,  which  is  —  1.      Therefore  the  Galois 

187 


188  THEORY   OF    EQUATIONS 

group  is  either  the  cyclic  group  of  the  degree  n  —  1  or  one  of 
its  sub-groups,  §  162.  Since  I  is  a  normal  equation,  it  is  its 
own  Galois  resolvent  ;  the  Galois  domain  is  of  the  degree  n  —  1 
and  the  Galois  group  of  the  order  ?i  —  1.  Hence  the  Galois 
group  of  I  is  the  cyclic  group  of  the  (n  —  l)th  order. 

Ex.  1.  If  n  is  prime,  show  that  xn  —  1  =  0  is  a  cyclic  equation  in  the 
domain  ii(i).  In  what  follows  we  shall  exclude  from  our  consideration 
cyclic  equations  whose  roots  are  not  all  irrational. 

171.  Theorem.  Each  root  of  acyclic  equation  can  be  expressed 
as  a  function  in  (i  of  any  other  root. 

If  a,  «!,••-,  «„_!  are  the  roots  of  the  cyclic  equation  f(x)  =  0, 
then  the  function  in  Q  of  x  of  the  (n  —  l)th  degree, 

*  (x)=  / 


—  a     x  — 


admits  the  permutations  of   the  cyclic  group  and  is,  there- 
fore,  a  number    in   O,    §    154.      If    we    put    in    succession 


x  —  a,  «j,  •••,  «„_!,  and  if  we  use  the  notation  —  ^  =  <f>(x),  we  get, 

J(x) 

§    142,         <*i  =  <£(«),    «s  =  <£(«,),   ••-,  «„_!  =  <K«»-2),  «  =   <K««-l)- 

This  holds  even  when/(a)  =  0  is  a  reducible  equation,  pro- 
vided that  it  has  no  multiple  roots. 

Ex.  1.  When  are  cyclic  equations  normal  ? 

Ex.  2.  Show  that  one  root  of  a  quadratic  equation  can  be  expressed 
as  a  function  in  0(a1(  aj  of  the  other  root. 

Ex.  3.  Show  that  any  root  of  a  cubic  can  be  expressed  as  a  function 
in  0(0!,  oj,  as,  ^D)  of  one  of  the  others. 

Ex.  4.  Show  that  «2  =  tf>2(«),  «5  =  «£8(«),  etc.,  where  the  superscript 
is  not  an  exponent,  but  indicates  that  the  functional  operation  <f>  is  to  be 
repeated.  Thus,  <f>2(a)  =<£(>(«)). 

Ex.  5.   Prove  that  «i  =  0n+1(«),  02  =  0M+2(«)>  etc. 


CYCLIC    EQUATIONS  189 


Ex.6.   If  0(a)  =  ««jL=aij    0s(a)  =  ««!_+    =ag    etc     then    it 

d 


may  be  shown  that  <t>m(a)  =  «,  when  a  +  d  —  2  cos  —  and  ad  —  be  =  1. 

m 

where   k  and  m  are   relatively   prime.     (See   Cole's  traiisl.  of   Netto's 
Theory  of  Substitutions,  pp.  204-207.)     Show  that  when  a  =  0,  —  b  =  c 

=  d  =  1,   k  =  1,   m  =  3,    we  have   ai  —  ---  ,    a»——l  —  -,   where 

«  +  1  a 

a,  «i,  «2  are  roots  of  the  cyclic  equation  x3  +  x2  —  2  x  —  1  =  0. 

Ex.  7.  Show  that  if,  in  Ex.  6,  a  =  0,  b  =  -c=d  =  k=l,  m  =  3, 
then  a,  al5  «2  are  roots  of  x3  +  aa;'2  —  (a  +  3)x  +  1=0. 

172.  Solution  of  Cyclic  Equations.  The  general  solution  of 
cyclic  equations  can  be  easily  obtained  by  the  aid  of  the 
Lagrangian  resolvents,  §  115. 

By  the  theorem  in  §  118  the  expression  represented  by 
[o>,  a]",  in  which  the  a,  al9  •••,  an_i  are  the  roots  of  f(x)  =  0, 
and  to  is  a  primitive  nth  root  of  unity,  §  66,  is  such  that  the 
coefficient  of  each  power  of  w  is  a  cyclic  function  of  the  roots 
of  f(x)  =  0.  See  Ex.  1,  §  119.  Thus  [w,  «]"  is  a  function  in 
fyoj,  02,  ...  an,  <o,)  which  belongs  to  the  cyclic  group.  This  function 
is  a  number  in  ty^,  <»2,...aB,  «)>  §  154.  Let  the  coefficients  of 
different  powers  of  a>  in  [a/,  «]"  be  c0,  GI,  •••,  cn_!.  Write 

[a/,  «]n  =  C0  +  Caa/  +  C2a>2A  H  -----  h  Cn_lW("-1)A  =  7\. 

The  cyclic  function  7\  can  be  computed.  Regarding  it  as 
known,  we  get  ^  a]  =  ^r^ 

Assign  to  X  the  successive  values  1,  2,  •••,  (n  —  1),  and  we  have 
a  +  wai     +  •••  +  w"-1^.! 


where  c^  is  known.     Adding,  we  get 

na  =  -  a,  +  VTl  +  ^/T.  +  -  + 


190  THEORY   OF   EQUATIONS 

I 

Thus  the  root  a  is  expressed  in  terms  of  radicals  of  the  ?ith% 
order,  where  the  2\  are  made  up  of  numbers  in  O(aii  O2...,a 
and  the  nth  roots  of  unity.     Each  of  the  radicals  in  /  has  n 
values  which  differ  from  each  other  by  a  factor  that  is  a  root  of 
unity. 

Our  expression  I  involves  a  difficulty  which  demands  our 
attention.  Since  each  radical  has  n  values,  it  follows  that  the 
(n  —  1)  radicals  represent  n"~l  values.  Hence  there  are  in  I, 
besides  the  n  roots  of  the  given  equation,  nn~l  —  n  foreign 
values,  and  no  method  is  assigned  for  telling  which  of  the 
values  represent  the  roots  of  the  given  equation. 

To  remove  this  difficulty,  H.  Weber  proceeds  as  follows  :  If 
we  effect  the  substitution  (0  1  2  •••  n—  1)  upon  [CD,  «]"~A  •  [o/,  «], 
then  by  §  119  the  indices  of  the  coefficients  of  this  product 
undergo  the  substitution  (0  1  2  •  •  •  (n  —  l))n~A+A.  As  this  is  the 
identical  substitution,  the  coefficients  are  unaltered. 

Let    [co,  «]-*  •  |>*,  «]  =  #A  =  e,/*>  +  £!<*>»  +  -.  +  e,,.,™*,"-1, 
then  E^  is  a  cyclic  function  in  fl(ai,  o2,  ...  an-1,  „>  and  may  be  con- 
sidered as  known.     We  have 


Hence  *T  *      =.  II 


From  II  it  appears  that,  for  a  fixed  primitive  value  of  w,  each 
of  the  radicals  which  appear  in  our  value  for  na  in  I  may  be 
expressed  as  a  function  in  Q  of  one  of  them.  If  that  one 
radical  be  given  all  its  n  values,  the  expression  for  na  has  n 
values  which  are  the  n  roots  of  the  given  equation. 

173.  Computation  of  TA.  In  most  cases  the  computation  of 
this  quantity  is  extremely  involved  and  special  devices  rrust 
be  resorted  to.  An  idea  of  such  devices  will  be  given  in  the 
discussion  of  cyclotomic  equations,  where  the  solution  is 
divided  up  into  the  simplest  component  operations.  We  give 
here  the  computation  of  Tl=(a-\-  a^  + 


CYCLIC    EQUATIONS  191 

A  =  «2«i  -f-  «!2«2  +  «22«, 

A'  =  «f« 


!  =  a3  +  «!3  +  «23  +  6  ««!«.,  +  3  uA  +  3  <JA' 


T2  =  1(9  dOs-2  a^-27  as)-f  V^TD^OS-SV 
where  £  =  9  avaz  —  2  af  —  27  a3.     We  have  now 


:  =  V-i(S  +  3V-3Z>). 

3  ; 


3  V  —  3  D). 

Having  thus  evaluated  the  Lagrangian  resolvents  for  the 
cubic,  we  can  readily  obtain  an  expression  for  the  roots  of 
the  general  cubic  by  adding  the  values  of  V^  and  -\/T2  to 
a  +  «!  +  «2  =  —  ai-  See  solution  of  Ex.  3,  §  162. 

Ex.  1.   For  the  quartic  x*  +  a,\x*  +  a2x2  +  a,%x  +  «4  =  0  compute 

T=  (a  +  w«i  +  a>%2  +  w8«3)4, 
where  w  =  i  or  —  i. 

Letting  TI  =  (a  +  *'«i  —  «2  —  las)*, 

Tn  =  (a  —  i«i  —  «2  +  i«3>4, 
we  have  TI  +  T2  =  2 (a  -  «2)4  -  12(«  -  «2)2(«i  -  "s)2  +  2(«i  -  «3)4 

=  4{(a  -  «2)2  -  («!  -  «3)2}2  -  2{(«  -  «2)2+  (ai-as)2}2 
=  4  pops  —  2  (of]2  —  2  «2  —  2  0i)2, 

where  0i  =  acta  +  ai«3  is  a  root  of  the  cubic  in  Ex.  11,  §  71, 
and  where        p2  =  (a  +  «i  -  «2  -  "s)2,  PS  =  («  -  «i  -  «2  +  «8)2. 
Let  pi  =  (a  —  «i  +  «2  —  «s)2, 

then  Pi  =  ai2  —  4  a2  +  4  fa,  pipzps  —  («i8  —  4  ai«2  +  8  as)2, 

Ex.  18,  §  71.     Hence  the  value  of  p2p3  is  known.     We  have  also 

7\r2=  (ffli2  -  2  a2  -  2  004. 
Hence  TI  and  Tg  are  roots  of  the  known  quadratic         , 


192  THEORY    OF   EQUATIONS 

Ex.  2.    Carry  out  the  computation  in  Ex.  1  by  taking 
ai  =  a2  =  0,  as  =  «4  =  5 

and  show  that    T  will  have  the  values    60  ±  80  i,   which    lie   in   the 
domain  $2(i,  <). 

Ex.  3.   Find  T\  and  T2  when  in  the  quartic  a\  =  a-2  =  04  =  0,  a3  =  1. 
In  this  case,  is  the  cyclic  group  the  Galois  group  in  fyi,<)  ? 

Ex.  4.   Taking          a  —  0.1  +  a?  —  «3  =  \/pi, 
a  +  «i  -  «2  -  «s  =  Vpz", 
a  -  «i  -  «2  +  «3  =  Vpij, 
give  a  solution  of  the  general  quartic,  pi,  p2,  PS,  being  roots  of 

P8  +  (8  a2  -  3  «!2)p2  +  (3  ai*  -  16  ai2a2  +  16  ctia3  +  16  a22  -  64  a4)p 
—  (ai3  -  4  a^s  +  8  «3)2  =  0.     See  Ex.  1. 

Ex.  5.   Find  a  solution  of  the  general  quartic  by  taking 
a  +  i«i  -  «2  -  i«3  =  \/?\, 

a  -  «i  +  «2  -  «s 

a  -  toi  -  «2  4-  i«s  =  - 

where  A  =  (a  —  «i  +  03  —  «s)  («  +  z«i  —  «2  —  i«s)~2 

+  (ai2  -  2  a2  -  2  ^Q2] 


2r1(4aia2-«i'5-8a3) 

£  =  (a  —  i«i  —  cca  +  ias)  (a  +  i«i  —  «2  —  iOs)~8 
_  ai2  -  2  a2  -  2  0! 
2i 

174.  Cyclic  Equations  of  Prime  Degree.  TJie  solution  of  any 
cyclic  equation  can  be  made  to  depend  upon  the  solution  of  cyclic 
equations  whose  degrees  are  prime. 

The  solution  in  §  172  applies  to  cyclic  equations  of  any  degree 
and  is  perfectly  general.  Nevertheless  it  is  of  importance,  for 
subsequent  developments,  to  prove  the  present  theorem.  We 
give  the  proof  for  the  degree  12  =  3  •  4.  The  generalization  to 
the  case  n  —  e  •  /  is  obvious. 


CYCLIC    EQUATIONS  193 

Let  s  =  (««!  -.  «„),  where  «x  =  <£(«),  «,  =  £(«,),  ^  =  <£(«.,),  ..., 
then  s3  can  be  resolved  into  three  cycles,  c,  c^  c2,  as  follows  : 


c  = 

Cl  =  («1«4«7«10), 

C  = 


Let  T/  be  a  function  $  in  O  of  the  roots  «,  «g,  og,  «,»  which 
belongs  to  the  cycle  c.  The  substitutions  of  the  Galois  group 
P=  {1,  s,  s2,  ...  s"-1]  of  /(oj)=0,  applied  to  y,  give  three  distinct 
values, 


2/1  =  <K«l«4«7«lo), 

y2  =  iK«2«5«8au), 
which  are  roots  of  a  cubic  equation, 


The  coefficients  of  Z  in  I  are  symmetric  functions  in  fi  of  y, 
2/i,  2/2,  and  are,  therefore,  unaltered  by  the  substitutions  of  P. 
Hence  these  coefficients  are  numbers  in  O,  §  154. 

We  proceed  to  show  that  I  is  a  cyclic  equation  whose  group 
is  PI=  {1,  (2/2/i2/2)>  (yy^/i)\-  Remembering  that  the  substitutions 
of  the  group  P  interchange  y,  yl}  y2  cyclically,  we  see,  firstly, 
that  any  function  of  y,  yi}  y2  which  admits  of  the  substitution 
of  P!  is  a  function  of  a,  al}  •••,  an_^  which  admits  of  the  substi- 
tutions of  P  (the  Galois  group  of  f(x)  =  0),  and  such  a  function 
is  a  number  in  O,  §  1  54  ;  secondly,  any  function  of  y,  yt,  y2, 
which  is  a  number  in  fi,  is  a  function  of  the  roots  «,  «u  •••,  «n_i« 
which  is  a  number  in  O  and  hence  admits  of  the  Galois  group  P, 
§  153,  thus  showing  that  the  function  of  y,  ?/„  y2  admits  of  the 
substitutions  of  Pv  Consequently  PI  is  the  Galois  group  of 
equation  I,  §  155. 
o 


194  THEORY   OP   EQUATIONS 

We  can  now  prove  that  f(x)  can  be  broken  up  into  three 
factors  of  the  fourth  degree  each,  thus, 

f(x)  =  F(x,  y)  -  F(x,  2/j)  -  F(x,  */2),  II 

where  F(x,  y)  =  0  is  a  quartic  cyclic  equation,  in  which  the 
coefficients  of  x  are  numbers  in  the  domain  I1(y).  For,  let 

FI(X)  =  (*  —  «)  (*  —  «u)  (»  —  On)  (*  —  «a),  HI 

then  each  coefficient  of  #  in  III  admits  the  circular  substitu- 
tion c;  hence  it  admits  also  the  substitutions  of  what  becomes 
the  Galois  group  of  f(x)  =  0  after  the  adjunction  of  y.  This 
group  must  consist  only  of  powers  of  c,  c1?  c2.  Therefore,  these 
coefficients  of  x  are  functions  of  ?/,  §  162,  and  we  have  F^x) 
=  F(x,  y).  Moreover,  F(x,  y~)  =  0  is  a  cyclic  equation  in  O(y), 
since  the  cyclic  functions  of  its  roots  lie  in  this  domain. 

If  in  n  =  e  •  f,  e  or/  are  composite  numbers,  then  we  repeat 
the  process  upon  the  new  cyclic  equations  until  all  the  factor 
equations  are  of  prime  degree. 

Thereby  the  solution  of  cyclic  equations  of  any  degree  n  is 
made  to  rest  on  the  solution  of  cyclic  equations  whose  degrees 
are  prime  numbers. 

Ex.  1.  As  an  illustration,  take  x4  -f  xs  +  x2  +  x  +  1  =  0,  where  a  =  w, 
ai  =  w2,  «2=w4>  a3  =  w8  =  w3.  Hence  s=(aai«2a3)  =  (wu2w*w3),  c  =  (wu>4), 
d  =  (w2w3).  Take  y  =  ««22  4-  «2«2  =  w4  +  w,  then  y\  =  a\az2  +  a3«i2 

=  or5  +  CO2,     y  +  yi  =  -  1,      ?/?/1    =  -  1,     («  -  y)(«  -  yi)  =  «*  +  «  -  1   =  0, 

2  «  =  _  1  ±  V5,  /(x)  =  (««  +  Q  -  J  V6)«  +  1)(<2  +  (J  +  i  V6)«  +  1)  = 
F(x,  «/)  •  -F(a;,  «/i).  Each  quadratic  factor,  equated  to  zero,  is  a  cyclic 
equation. 

Ex.  2.  Given  that  f(x)  =  x6  +  x5-5x4-4a;8  +  6x2  +  3a;  —  1=0  is 
a  cyclic  equation  in  which  a  =  2  cos  a,  «i  =  2  cos  na,  «2  =  2  cos  w2a,  •••, 

o  _ 

«5  =  2  cos  n6a,  where  «  =  2  and  a  =  — ;-  •     In  illustration  of  the  theorem, 

13 

we  have  s  =  (aai«2«3«4«5),  c=(a«o«4),  Ci=(aia3«6).  Take  y  =  ««22 
-f  a2«42  +  «4«2,  yi  =  «i«32  4-  «3«52  +  «s«i2.  With  some  effort  we  find 
y+2/!=-5,  yyi=3.  Hence  («-2/)(«-yi)=«2  +  5  «+3=0,  2f=-5±Vl3. 
We  get  /(x)  =  (*3  -  df2  -  f  +  d  -  1)  («3  +  (d  +  l)f2  - 1  -  d  -  2)  =  0,  where 
2  d  =  -  1  ±  VI3. 


CYCLIC    EQUATIONS  195 

The  cubic  factors  yield  cyclic  equations  of  prime  degree.  The  expres- 
sion for  y,  selected  in  this  example,  is  somewhat  unwieldy.  A  better 
choice  is  made  in  the  periods  of  §  180. 

Ex.  3.  If  m  is  odd,  and  equal  to  2  n  +  1,  show  that  (gm  ~  *)  =  Q, 
when  z  +  -  =  xt  yields  the  cyclic  equation 

|p 

0  =  a?>+  jf-1  -(»  -  l)z»-2  -(u  -  2)z''-3  |  ~ 


J,  *  2s 

1        - 


1.2 

which  has  the  roots  a  =  2  cos  £a,  where  a  =  —  2-^—  ,  and  where  k  takes 

2  n  +  1 

successively  the  values  1,  2,  3,  ••-,  n.     When  2  n  +  1  is  prime,  the  equa- 
tion is  irreducible. 

175.   Theorem.    Every  function  in  O  of  the  roots  of  an  irreduci- 
ble cyclic  equation  is  itself  the  root  of  a  cyclic  equation. 

Let  a  be  a  root  of  the  given  irreducible  cyclic  equation  and 
g(a)  the  function.     Then  if  the  values 


gr(a), 

are  not  all  distinct,  let  say  g(a)  =  #(<£*(«)),  and  we  have,  §  138, 
the  rectangle 

g(a), 


in  which  the  values  in  each  column  are  equal,  while  the  values 
in  each  row  are  distinct,  and  are  roots  of  an  irreducible  equa- 
tion in  O,  viz., 


h(y)  =  (y-  g(a))(y  -  0(*(a)))  -  (y  -  ^(^(a)))  =  0. 
The  consideration,  as  in  §  142,  of  the  function 


196  THEORY    OF   EQUATIONS 

leads  to  the  conclusion  that 


A  similar  conclusion  is  reached  if  all  the  values  of  I  are 
distinct. 

Ex.  1.  If  w  is  a  complex  fifth  root  of  unity,  show  that  1  +  o>,  1  +  w2, 
1  4-  w3,  1  +  w*  are  roots  of  a  cyclic  equation. 

Ex.  2.   By  §  175  form  the  roots  of  a  cyclic  equation  of  the  sixth  degree. 

Ex.  3.  Show  that  in  a  domain  made  up  of  real  numbers  :  (1)  a  cyclic 
equation  has  all  its  roots  real,  if  one  is  real  ;  (2)  all  the  roots  of  a  cyclic 
equation  of  odd  degree  are  real  ;  (3)  all  the  roots  of  a  cyclic  equation  of 
even  degree  are  complex  when  one  of  them  is  complex. 

176.  General  Cyclic  Cubic  Equation.  To  determine  the  general 
irreducible  cyclic  equation  of  the  third  degree,  let  «,  al}  a.,  be 
the  roots  of  the  required  cubic,  where  «1  =  <^>(«),  a2  =  <f>(al). 
From  §  80,  it  follows  that  the  most  general  algebraic  function 


By  §  175,  da  -f  e  is  also  a  root  of  a  cyclic  equation.  Writ- 
ing da  +  e  for  a  in  I  and  selecting  for  d  and  e  values  which 
cause  the  coefficient  of  a  to  disappear  and  that  of  a2  to  be 
unity,  we  obtain  a  simpler,  yet  general  function,  <£  (a)  =  a2  +  c. 

Wehave  ai  =  «2+c, 

«2  =  «!2  +  C, 

a  =  of  -}-  c. 
Eliminating  «i  and  «2,  we  have 

(a2  +  c)4  +  2  c(«2  +  c)2  -  a  +  c2  +  c  =  0. 

Since  «i  cannot  equal  «,  the  expression  «j  —  a  =  (a2  +  c)  —  a 
cannot  be  zero.     Dividing  by  (a2  +  c)  —  a,  we  get 
c  +  ^a4  +  (2  c  +  i)«s  +  (3  fS  +  3  c 

=  0. 


CYCLIC    EQUATIONS  197 

If  the  required  cubic  is  of5  —  a-^  +  a&  —  as  =  0,  then 
(*!  =  a  +  «!  +  oj,  =  «4  +  (2  c  +  1)«2  +  a  +  (c2  +  2  c), 
02  =  a6  +  a5  +  3  c«4+  (2  c+l)«3+  (3  c2+c)«2+(c2+2  c)a+  (c3+c2). 

Byll,  =_ai  +  (c_i). 

a3  =  a7  +  3  ca5  +  (3  c2  +  c)cf  +  (c3  -f  c*)a. 

By  II,  =ca1  + 


Equation  II  is  satisfied  by  the  three  roots  a,  «„  «._,  and  also  by 
three  other  roots  «',  u\,  a'^  whose  sum  we  designate  by  a  \. 
We  have 


aia\  =  3  c  +  1  +  cti  +  a\  —  2(c  —  1), 

=  c  +  2, 
and  o^  a\  are  roots  of  the  quadratic 


Since  the  sextic  II  is  satisfied  by  the  roots  a,  «j,  «.,  of  the  irre- 
ducible cubic,  II  must  be  reducible  into  two  cubics.  Hence  at  and 
a\  must  be  numbers  in  £2.  Hence  the  discriminant  —  (4  c  +  7) 
of  the  quadratic  must  be  a  perfect  square  ;  in  other  words, 


or  c  =  _(/>+/+  2). 

The  roots  of  the  quadratic  are  /and  —  (/-f  1).  Writing  al=f. 
we  get  a2  =  -(/2+2/+3),  a3=(/3+2/2  +  3/+l).  Thus  the 
coefficients  of  the  required  cubic  are  obtained,  where  /  is  any 
number  in  fl.  To  remove  the  second  term  of  this  cubic,  take 

=  a;-,  and  we  get 


f  _  3(m2  +  m  4.  i)y  4.  (m«  +  m  +  1)(2  m  +  1)  =  0.        Ill 

Every  cyclic  equation  of  the  third  degree  can  be  reduced  to  III. 
See  Ex.  4,  §  159. 


198  THEORY   OF   EQUATIONS 

Ex.  1.    Show  that  the  discriminant  of  III  is  a  perfect  square, 

D  =  92(m2  +  m  +  I)2. 
Ex.  2.    For  the  equation  III  determine  the  function  <t>  in  the  relation 

«i  =  <£(«)• 

Ex.  3.  Any  cyclic  equation  of  the  fourth  degree  can  be  reduced  to  the 
form  yi-2  6(2  s  +  r2)y2  -  4  6r(l  +  6s2)y  +  62(r2  -2s)2-  6(1  +  6s2)2  =  0, 
where  6,  r,  s,  are  rational  numbers  and  6  is  not  a  perfect  fourth  power. 
See  Ex.  11,  §  159.  Prove  that  this  equation  can  be  solved  without  the 
extraction  of  cube  roots. 

CYCLOTOMIC  EQUATIONS;   GEOMETRIC   CONSTRUCTIONS 

177.  Introduction.  In  §  63  and  §  64  it  was  shown  that  the 
roots  of  x"  —  1  =  0  may  be  represented  thus, 


where  k  takes  successively  the  values  0,  1,  •  •  •,  n  —  1,  and  that 
the  solution  of  af  —  1  =  0  is  geometrically  equivalent  to  the 
division  of  the  circumference  of  a  circle  into  n  equal  parts. 
The  solution  of  xn  —  1  =  0,  given  in  §  63,  is  trigonometric.  We 
proceed  to  show  that  it  is  always  possible  to  give  an  algebraic 
solution.  We  shall  point  out  how  this  solution  can  be  effected 
and  shall  consider  the  cases  in  which  the  division  of  the  circle 
into  equal  parts  can  be  effected  with  the  aid  of  the  ruler  and 
compasses. 

178.   Cyclotomic  Equations.     If  we  remove  the  root  1  from 
a?"  —  1  =  0  by  dividing  by  x  —  1,  we  obtain 

=  0.  I 


If  n  is  a  prime  number,  equation  I  is  called  a  cyclotomic  equation. 
In  the  domain  O(1)  the  cyclotomic  equation  is  irreducible,  §  130, 
and  cyclic,  §  170. 

If  n  is  a  composite  number,  we  know  from  §  66  that  the  solu- 
tion of  of1  —  1  =  0  can  be  reduced  to  the  solution  of  binomial 


CYCLIC   EQUATIONS  199 

equations  of  the  form  xm  —  A  =  0,  in  which  the  exponents  ra  are 
the  prime  factors  of  n.  By  taking  x*\/A  =  z,  the  equation 
xm  —  A  =  0  becomes  zm  —  1  =  0.  Hence  the  general  solutions  of 
binomial  equations  can  be  given  as  soon  as  we  are  able  to  solve 
binomial  equations  of  the  form  zm  —  1  =  0  whose  degrees  are 
prime  numbers.  It  is  the  latter  equations  which  by  division 
by  z  —  1  give  rise  to  the  cyclotomic  equations. 

Since  a  cyclotomic  equation  is  a  cyclic  equation,  its  solution 
is  theoretically  contained  in  §  172.  But,  as  a  rule,  the  compu- 
tation of  2\  is  extremely  involved.  We  proceed  to  develop  a 
scheme,  due  to  Gauss,  by  which  the  solution  of  cyclotomic 
equations  is  divided  into  simpler  component  operations. 

Ex.  1;   Show  that  cyclotomic  equations  are  reciprocal  equations. 

179.  Primitive  Congruence  Roots.  It  is  shown  in  the  Theory 
of  Numbers  that,  for  every  prime  number  n,  there  exist  num- 
bers g  (called  primitive  congruence  roots  of  n),  such  that,  on 
dividing  by  n  each  member  in  the  series, 

9,  92,  9s,  '",  9n~\ 

the  remainders  obtained  are  (except  in  their  sequence)  the 

numbers  in  the  series  _ 

1,  2,  3,  -..,  n-l. 

For  instance,  if  n  =  5,  we  may  take  g  =  2.  If  2,  22,  23,  24  are 
each  divided  by  5,  the  remainders  are  respectively  2,  4,  3, 1. 
These  remainders  differ  from  the  series  1,  2,  3,  4  only  in  the 
order  in  which  they  come.  Illustrate  the  same  by  taking  n  =  7 
and  g  =  3. 

In  view  of  these  facts  and  of  the  relation  <on  =  1,  the  roots 
o),  wj,  •••,  a>rt_!  of  the  cyclotomic  equation  I  may  be  written  thus: 
CD  =  w,  a)!  =  a/,  w2  =  a/,  •••,  a>n^2  =  a)"""2.  This  notation  will 
offer  certain  advantages.  The  roots  of  I  may  therefore  be 
written :  __« 


200  THEORY    OF    EQUATIONS 

Ex.  1.  By  trial  find  the  smallest  integer  that  may  be  taken  as  the  value 
for  g  when  n  =  11,  and  show  that  to,  w^,  w»2,  •••,  us10  represent  the  same 
roots  as  w,  w2,  w3,  •••,  w10.  Show  that,  for  n  =  13,  g  may  be  2  or  6. 

180.  Solution  of  Cyclotomic  Equations  reduced  to  Equations  of 
Prime  Degree.  As  is  evident  from  §  174  we  can  base  the  solu- 
tion of  equation  I  of  §  178  upon  cyclic  equations  whose  degrees 
are  prime  factors  of  w—  1.  When  n  is  prime,  n  —  1  is  composite. 
Let  n  —  1  =  e  •  f,  where  e  is  a  prime  factor.  As  before,  let  w  be  a 
root  of  the  cyclotomic  equation  I.  Then  construct  expressions 
i])  "ni>  '  '  ')  ^e-u  called  periods,  as  follows  : 


-f  o/2e  -\ 


Ill 


In  each  period  there  are  /terms  and  the  first  term  is  the  #eth 
power  of  the  last  term,  and  each  of  the  terms  after  the  first  is 
the  gth  power  of  the  term  preceding  it.  Each  of  the  periods 
is,  therefore,  a  function  that  belongs  to  the  cyclic  group 


where  the  substitution  s=  (w,  u>l}  w2,  •••,  wn_2)-  The  periods  III 
are  special  forms  which  the  functions  y,  ylt  y2  in  §  174  may 
assume.  From  §  174  it  follows  that  the  periods  III  are  the 
roots  of  an  irreducible  cyclic  equation 

(aj-7)(aj-7l)...(aj-7._1)=0.  IV 

This  is  an  equation  in  O  and  of  th6  degree  e.  By  the  solution 
of  this  equation  the  periods  become  known  quantities. 

181.  Product  of  Two  Periods.  In  order  to  compute  the  co- 
efficients of  equation  IV  in  §  180  we  must  multiply  periods 
one  by  another.  Take 


CYCLIC    EQUATIONS  201 

Observing  that  yh  remains  unaltered  when  <•/  is  replaced  by 
or  by  any  of  the  other  roots  in  that  period,  we  may  write 
the  product  of  the  two  periods  as  follows  : 


+  a/*"-1  V*+(/~ 
In  this  product  the  terms  in  the  first  column  are, 


If  (gk  +  gh)  is  a  multiple  of  n,  then  this  column  becomes 
equal  to/.  If  (g^  +  gf*)  is  not  a  multiple  of  n,  then  this  column 
is  one  of  the  periods  in  III,  §  180. 

The  same  conclusion  is  reached  for  every  column  in  the 
product.  Hence  the  product  is  a  linear  function  of  the  periods, 
the  coefficients  in  this  function  being  numbers  in  the  given 
domain  12(1). 

182.  When  f  is  a  Composite  Number.  When  in  the  relation 
n  —  1  =  e  •  /,  both  e  and  /  are  prime  numbers,  the  solution 
of  the  cyclotomic  equation  is  evidently  made  to  depend  on  the 
solution  of  two  equations  whose  degrees  are  prime,  one  equa- 
tion being  of  the  degree  e,  the  other  of  the  degree  /. 

When  /is  a  composite  number,  one  or  more  additional  steps 
are  necessary  to  reduce  the  problem  to  the  solution  of  equa- 
tions of  prime  degree.  If  /=  e'  «/',  where  e'  is  prime,  we  may 
form  ee'  periods,  with/'  terms  in  each,  as  follows  : 


t  a  a"'+l       i          o2"^1      I  I         o< 

w  i  =  of  +  or          +  of  +  ' '  *  +  or 

,  J2e  0««'+2«     .        -2«'+2«  .        „( 

ly'a,  =  wT  +0^          +a/          H \-  uf 


202  THEORY   OF    EQUATIONS 

It  is  to  be  noticed  that,  if  we  select  every  eth  period  in  this 
set,  the  sum  of  the  periods  thus  selected  is  equal  to  one  of  the 
known  periods  III,  §  180.  For  instance, 


These  periods  if,  rj'e,  7/'2(.,  •••  are  roots  of  an  irreducible  cyclic 
equation  of  the  degree  e',  the  coefficients  of  which  are  linear  func- 
tions of  the  known  periods  III. 

If  f  is  a  composite  number,  repeat  the  above  process  by 
assuming  f'  =  e"-f".  If  n  =  e  •  e'  •  e"  •  /",  then  the  above 
process  calls  for  the  solution  of  one  equation  of  each  of  the 
prime  degrees  e,  e',  e",f".  As  soon  as  one  root  of  a  cyclotomic 
equation  is  found,  the  others  can  be  obtained  by  raising  that 
one  to  the  2d,  3d,  •••,  nth  powers. 

183.  Constructions  by  Ruler  and  Compasses.  The  operations 
of  addition,  subtraction,  multiplication,  and  division  can  be 
performed  geometrically  upon  two  lines  of  given  length.  For 
instance,  in  elementary  geometry  we  learn  how  to  construct 
the  quotient  of  a  line  a  inches  long  and  another  line  b  inches 
long,  by  the  aid  of  the  proportion  x  :  1  =  a  :  b.  In  elementary 
geometry  we  learn  also  how  to  construct,  by  means  of  ruler 
and  compasses,  the  irrational  Vab.  The  geometric  construction 
of  v  c  +  Va&  is  simply  a  more  involved  application  of  the  pro- 
cesses just  referred  to.  But  we  are  not  able  to  construct  with 
ruler  and  compasses,  irrationals  like  ^Jab.  Thus  it  is  evident 
that  all  rational  operations  and  those  irrational  operations 
which  involve  only  square  roots  can  be  constructed  geometri- 
cally by  the  aid  of  the  ruler  and  compasses. 

Conversely,  any  geometrical  construction  which  involves  the 
intersection  of  straight  lines  with  each  other  or  with  circles, 
or  the  intersection  of  circles  with  one  another,  is  equivalent  to 
rational  algebraic  operations  or  the  extraction  of  square  roots. 
This  is  the  more  evident,  if  we  remember  that  analytically 
each  line  and  circle  used  in  the  construction  is  represented  by 


CYCLIC   EQUATIONS  203 

an  equation  of  the  first  degree  and  second  degree.  Hence  there 
is  a  one-to-one  correspondence  between  constructions  by  ruler 
and  compasses  and  algebraic  operations  which  are  purely  rational 
or  involve  square  roots. 

Consequently,  if  we  wish  to  show  the  impossibility  of  con- 
structing a  quantity  by  ruler  and  compasses,  we  need  only 
show  that  the  algebraic  expression  for  that  quantity  in  terms 
of  the  known  quantities  cannot  be  given  by  a  finite  number  of 
square  roots. 

Applying  these  ideas  to  the  problem  of  dividing  the  circle 
into  n  equal  parts  by  means  of  ruler  and  compasses,  the  prob- 
lem is  possible  or  impossible  according  as  the  roots  of  x"  —  1  =  0 
can  be  expressed  by  a  finite  number  of  square  roots  or  not. 

If  n  is  a  prime  number  of  the  form  2*  +  1,  the  degree  n  —  1 
of  the  cyclotomic  equation  is  a  power  of  2,  and  the  operations 
called  for  in  §  182  involve  square  roots  only.  Hence,  when  n 
is  a  prime  of  the  form  2*  + 1,  the  division  of  the  circle  into  n 
equal  parts  by  ruler  and  compasses  is  ahvays  possible.  This 
important  result  is  due  to  Gauss. 

Ex.  1.    Solve  x5  —  1  =  0  by  Gauss's  method. 

The  cyclotomic  equation  is  x*+x3+x2+x+l=Q.  Here  n— 1=4=2-2; 
e  =  2,  /  =  2.  It  is  only  necessary  to  solve  two  quadratics.  By-  trial  we 
get  for  n  —  5,  g  —  2  the  roots 

w,  w»,  «»2,  0*9°; 
these  yield  the  two  periods 

i)  =  w  +  us*  =  «  +  «*, 
i7t  =  w9  +  w?3  =  w2  +  o»8. 

Hence  equation  IV,  §  180,  becomes 

z2  -  0?  +  *?i)*  +  im  =  0. 

But  17  +  i?i  =  w  +  w2  +  w3  +  w4  =  —  1, 

and  17171  =(w  +  w*)O2  +  w3)=  w8  +  w2  +  w  +  w4  =  - 1. 

Hence  the  quadratic  takes  the  form 

X2  +  X-1=0,  and  x  =  ~ 


204  THEORY   OF   EQUATIONS 

Take  17  =  —    -  --    The  quadratic  whose  roots  are  «  and  w4  is 
2i 

X2  —  (w  +  w*)x  +  to  •  w*  =  0, 

or  x2  —  t]X  +  1  =  0. 


Whence  g  =? 
2 

According  to  §  183  the  inscription  of  a  regular  pentagon  into  a  circle 
can  be  effected  with  the  aid  of  ruler  and  compasses. 

Ex.  2.    Solve  x13  -  1  =  0. 

Here  n  —  1  —  3  •  2  •  2.  Hence  the  solution  of  one  cubic  and  two  quad- 
ratics is  called  for,  and  the  inscription  of  a  regular  polygon  of  thirteen 
sides  into  a  circle  by  ruler  and  compasses  is  impossible.  Take  g  =  6,  then 

the  roots  of  -  —  —  •  =  0  are 
x  —  1 

W,     <j)9,     W»2,     •••,     Wf", 
Of  W,     W6,     O)10,     W8,     W9,     W2,     W12,     W7,     W8,     W5,     W*,     W1*. 

If  we  take  n  —  !  =  «•/=  12  =:3-  4,  where  e  =  3,  we  get 
17  =  w  +  w8  +  w12  +  w5, 

7/1  =  W6  +  W9  +   W7  +   W*, 
1J2=W10+  W2+  W3+  W11. 

To  compute  the  cubic  of  which  17,  rji,  ?;2  are  roots,  we  obtain 


77171  =  2  17  +  771  -f  772, 

^i^  =  i?  +  2  171  +  172, 

Viz  =  V  +  ill  +  %  "02, 

1777  =  4  +  2  tji  +  172, 

1777x772  =  1717  +  2  1777!  +  77772  =  —  1, 

Wi  +  171172  +  W2  =  4(77  +  77!  +  772)  =  —  4. 

The  cubic  is  x8  +  x2  —  4  x  +  1  =  0.     Solving  this,  we  obtain  the  values  of 

i?)  i?i,  172- 

Take  next  /  =  4  =  e'f  =  2  •  2.     We  have  77'  =  w  +  w12,  TJ'S  =  w8  +  w8. 
Since  17'  +  rj'a  =  77  and  77V3  =  771,  we  find  that  77'  and  77*3  are  roots  of  the 

quadratic 

Xs  -  77X  +  77!  =  0, 


CYCLIC   EQUATIONS  205 

and  are  therefore  known.     Next  form  the  quadratic  whose  roots  are  w 
and  w12.     Since  w  +  w12  =  77'  and  w  •  w12  =  1,  this  quadratic  is 

X2  -  r?'x  +  1  =  0. 

Either  root  of  this  quadratic  is  a  primitive  root  of  the  cyclotomic  equation, 
from  which  all  the  other  roots  may  be  found. 

Ex.  3.    Solve  x17  -  1  =  0. 

One  root  is  1.  To  find  one  of  the  primitive  roots,  form  the  cyclotomic 
equation  of  the  16th  degree  and  take  g  —  3.  Then  the  roots  are  repre- 
sented by  the  following  powers  of  w : 

i)  9i  #2>  yzi  (?•>  '"i  yl5i 

which  are  equivalent,  respectively,  to  the  powers 

1,  3,  9,  10,  13,  5,  15,  11,  16,  14,  8,  7,  4,  12,  2,  6. 
Take  n  -  1  =  16  =  e  •  /=  2  •  8,  where  e  =  2.    Then 

77   =  w  +  O)9  +  w13  +  u>15  4-  w16  +  w8  4-  w4  4-  o>2, 

77!  =  W3  4-  W10  +   W5  +  WU  +  WU  +   W7  +  W12  4-  W6. 

We  find  that  77  4-  771  is  equal  to  the  sum  of  all  the  roots,  or  —  1,  while 
77771  =  —  4.     Hence  77  and  771  are  roots  of 

z2  +  x  -  4  =  0. 
Next  we  take  /=  8  =  e'f  =  2-4,  where  e'  =  2  ;  then 

77'  =  w  4-  w18  +  w16  +  w4, 
ij'i  =  w8  4-  w5  4-  w14  4-  w12, 
77'2  =  w9  4-  <»15  4-  w8  4-  w2, 
77'3  =  w10  4-  w11  4-  w7  4-  w«. 
The  periods  77'  and  7j'2,  whose  sum  is  77,  are  roots  of 

X2  -  T7X  -  1  =  0, 

while  77'!  and  7j'3,  whose  sum  is  771,  are  the  roots  of 
x2  -  i?ix  -  1  =  0. 

We  get 


206  THEORY   OF   EQUATIONS 

In  the  third  step,  /'  =  4  =  e"f"  =  2-2, 

1)'       =  W  +  W16,  1/'4  =  Wl»  +  «*, 

Vi  =  w3  +  ww,  i/'5  =  w5  +  wl2, 

T7»2  =  W9  +  W8,  1j"6  =  W15  +   W2, 

1/'3  =  «W  +  W7,  ,//7  =  wll  +  uB. 

Since  tj"  and  i/'4  have  if  for  their  sum  and  ri'i  for  their  product,  they  are 
the  roots  of 

x2  -  -n'x  +  -n\  =  0, 

and  we  obtain 

Finally  we  find  that  w  and  w16  are  roots  of  the  quadratic 
7?  -  r,"x  +  1  =  0; 


//  \~lt-2 


„  ~ 

that  is,  o  =  1-  +  <-  -  1, 

2        ^4 

a  primitive  root  of  the  cyclotomic  equation  of  degree  16. 

After  solving  one  of  the  quadratics  given  above,  the  question  arises, 
which  one  of  the  two  roots  represents  a  given  period  ?  For  instance, 
which  of  the  roots  of  *2  —  171*  —  1  =  0  represents  T?'I  ?  To  settle  this, 
form  the  product 


Hence  Vi  —  1/3  is  positive,  and  TJ'I  has  the  plus  sign  before  its  radical, 
Vs  the  negative  sign. 

It  is  readily  seen  that,  since  the  equation  x17  —  1  =  0  involves  in  its 
solution  no  other  irrationals  than  square  roots,  a  regular  polygon  of  seven- 
teen sides  can  be  inscribed  in  a  circle  by  means  of  the  ruler  and  compasses. 
Gauss  discovered  a  method  of  inscribing  this  polygon  when  he  was  a  youth 
of  nineteen  years.  It  was  this  discovery  which  induced  him  to  pursue 
mathematics  as  his  life-work  rather  than  languages.  For  an  explanation 
of  the  construction  of  the  regular  seventeen-sided  polygon  consult  Bach- 
mann,  Lehre  von  der  Kreistheilung,  Leipzig,  1872,  p.  67,  or  Klein's  Famous 
Problems  of  Elementary  Geometry  (ed.  W.  W.  Beman  and  D.  E.  Smith), 
Boston,  1897,  p.  41.  We  have  followed  Bachmann's  exposition  of  the 
subject  of  the  division  of  the  circle. 


CYCLIC    EQUATIONS  207 

Ex.  4.  Show  the  impossibility  of  constructing,  with  ruler  and  com- 
passes, the  side  of  a  cube,  the  volume  of  which  is  twice  the  volume  of  a 
given  cube. 

(To  construct  a  cube  whose  volume  shall  be  double  that  of  a  given  cube 
is  the  problem  known  as  the  "  Duplication  of  the  Cube."  It  was  one  of 
three  problems  upon  which  Greek  mathematicians  expended  much  effort. 
Myth  ascribes  to  it  the  following  origin  :  The  Delians  were  suffering  from 
a  pestilence  and  were  ordered  by  the  oracle  to  double  a  certain  cubical 
altar.  Thoughtless  workmen  constructed  a  cube  with  edges  twice  as  long. 
But  brainless  work  like  that  did  not  pacify  the  gods.  The  error  being 
discovered,  Plato  was  consulted  on  this  "Delian  problem."  Through 
him  it  received  the  attention  of  mathematicians.) 

Ex.  5.  Show  the  impossibility  of  trisecting  by  the  aid  of  ruler"  and 
compasses  any  given  angle. 

To  trisect  a  given  angle  is  the  second  of  the  three  famous  problems 
first  studied  by  Greek  mathematicians.  The  third  was  the  "  Quadrature 
of  the  Circle." 

Let  x  be  a  complex  number  OA'  of  unit  length.    Let 

\AOB  -  0,    \AOA'  =  \A'OA"  =  \A"OB  =  * 


•  - 

o 

<t>         .    0 
Then  x  =  cos  ^  +  i  sm  •£, 

o  o 

20  ,       .20 
3?  =  cos  -£-  +  i  sm  -^-» 

o  o 

and  y?  =  cos  0  +  i  sin  0.  I 

According  to  our  problem  we  are  given  I,  where  xs  =  OB,  and  we  are  to 
show  the  impossibility  of  constructing  OA'  by  ruler  and  compasses. 

We  are  going  to  prove  that  equation  I,  as  a  rule,  is  irreducible.  It  is 
sometimes  reducible.  For  instance,  when  0  =  90°,  equation  I  gives  x*=i. 
which  can  be  factored  into  (x+i)(z2  —  ix  —  1),  which  factors  are  functions 
in  Od^).  In  this  case  the  construction  can  be  effected. 

When  the  right  member  of  I  is  an  arbitrary  number,  that  is,  when  0  is 
an  arbitrary  angle,  then  I  is  irreducible,  else  at  least  one  of  its  roots  could 
be  represented  as  a  function  of  cos  0  and  sin  0.  By  De  Moivre's  Theorem 


the  roots  of  I  are 


0          .     0 
5  +  tsm  5, 

o  o 

0  +  2ir  . 

—  -  +  z  sin 


0  +  4lT     .      .     •        0+4T 

=  cos  Z-21  —  f  i  sm  r_E_  —  . 


208  THEORY   OF   EQUATIONS 

If  in  these  expressions  for  x\,  x%,  Xa  we  substitute  </>  -f  2  IT  for  0,  the 
roots  undergo  a  cyclic  permutation  ;  that  is,  x\  becomes  x%,  x2  becomes  xs, 
Because  of  these  changes,  no  root  can,  in  general,  be  a 
rational  function  of  sin  0  and  cos  0  ;  for,  sin  <j> 
and  cos  0  remaining  unaltered  in  value  when 
<t>  +  2  w  is  substituted  for  0,  the  root  could 
undergo  no  change.  For  an  arbitrary  angle 
the  equation  I  is,  therefore,  irreducible.  Its 
degree  being  3,  which  is  not  an  integral  power 
of  2,  its  roots  cannot  be  constructed  with  the 
aid  of  the  ruler  and  compasses,  and  the  trisec- 
tion  is  impossible. 

Ex.  6.    Show  that,  if  we  take  cos  —  equal  to  a  value  a,  numerically  <|  1 

3 

and  rational  or  involving  square  roots  only,  we  get  x3  =  (a  +  i£)3,  where 
/S2  =  1  —  a2,  and  where  x  =  a  +  iff  is  a  root  which  can  be  constructed 
geometrically.  Show  that  any  number  of  trisectable  angles  0  may  be 
obtained  by  this  process.  Taking  a  =  £  -\/2  —  V3,  snow  that  the  angle 
of  45°  may  be  trisected.  By  assuming  «  to  involve  at  least  one  radical 
whose  order  is  not  two  nor  a  power  of  two,  show  how  to  obtain  angles 
which  cannot  be  trisected. 

Ex.  7.   Assuming  2  cos  —  =  a;,  show  that  the  trisection  of  the  angle  <f> 
3 

depends  upon  the  equation  xs  —  3  x  =  2  cos  0.  Letting  cos  0  =  m/n  and 
nx  =  y,  derive  ys  —  3  n2y  =  2  ?wn2,  which  has  integral  roots  whenever  the 
first  cubic  has  rational  roots.  If  the  integers  m  and  n  are  prime  to  each 
other,  and  n  is  divisible  by  an  odd  prime  p  but  not  by  p2.  show  that  0 
cannot  be  trisected.  Prove  that  angles  120°,  60°,  30°,  cos-1  £  cannot  be 
trisected. 

Ex.  8.  To  show  that  an  irreducible  cubic,  whose  coefficients  are  rational 
numbers  and  whose  three  roots  are  real,  cannot  be  solved  by  real  radicals. 

This  is  the  so-called  "  irreducible  case,"  §  60.  We  are  required  to  prove 
that  in  the  algebraic  solution  of  the  given  cubic  it  is  impossible  to  avoid 
the  extraction  of  the  cube  root  of  a  complex  number.  To  this  end  observe, 
first  (§  171,  Ex.  3)  that  the  cubic  becomes  a  normal  equation  when  VJ) 
is  adjoined  to  ft.  Here  VZ)  is  real.  The  equation  z"  —  a  =  0,  where  a  is 
not  a  perfect  nth  power,  and  n  is  prime,  is  irreducible.  If  it  were  possible 
for  the  normal  cubic  equation  to  become  reducible  on  the  adjunction  of 
the  real  root  X=  v/a,  then  by  §  166,  Cor.  II,  the  degree  of  xn  —  a  =  0 
would  be  a  multiple  of  j,  the  index  of  the  new  Galois  group  P=  1,  under 


CYCLIC    EQUATIONS  209 

GaW.  Here  this  index  is  3.  As  n  is  prime,  w=3.  This  makes  fyj-^Q^), 
where  p  is  a  root  of  the  normal  cubic.  Hence  the  roots  of  xn  —  a  =  0  are 
the  conjugate  values  of  JT,  §  136,  and  all  of  them  lie  in  the  normal  domain 
O(p>.  Now,  if  one  root  of  a  normal  equation  is  real,  all  its  roots  are  real. 
Therefore,  all  the  roots  of  x"  —  a  =  0,  being  functions  in  O  of  p  would 
have  to  be  real.  But  this  cannot  be,  when  n  —  3.  Thus,  the  assumption 
that  our  cubic  can  be  solved  by  real  radicals  of  prime  order  leads  to  an 
absurdity. 

Nor  would  the  solution  be  possible  by  real  radicals  of  composite  order, 
such  as  -v/a,  where  n  =pq,  &  composite  number  ;  for,  in  that  case  we  .can 
write  \'  -tya  and  we  can  adjoin  in  succession  the  radicals  of  prime  order 
y=  Va  and  Vy.  But,  as  has  just  been  shown,  such  adjunctions  do  not 
render  the  normal  cubic  reducible. 


CHAPTER   XVIII 

ABELIAN  EQUATIONS 

184.  Definition.  An  equation  f(x)  —  0  of  the  nih  degree, 
having  the  roots  a,  «„  •  ••,  «n_i  is  called  Abelian,  if  each  root 
can  be  expressed  as  a  function  in  O  of  some  one  of  its  roots, 

«1  =  <M«)>     «*=&(«)>     '"I     «»-!  =  <£n-l(«)> 

and  if,  for  any  two  of  these  roots,  we  have  the  commutative 
relation 

<£A<M«)  =  <£*<£A(«)- 

By  tf)h<f>k(a)  we  mean  here  <£A[<£A(a)]. 

The  equation  x4  —  1  =  0  is  Abelian,  because,  its  roots  being 
±  1,  ±  i,  we  have  —  1  =  t2,  —  i  =  i3,  1  =  i4,  (i2)3  =  (i3)2,  etc. 

Ex.  1.    Show  that  cyclic  equations  are  special  cases  of  Abelian  equations. 

Ex.  2.  Show  that  z6  —  1  =  0  is  Abelian,  but  not  cyclic  ;  that  z3  —  1  =  0 
is  both  Abelian  and  cyclic. 

Ex.  3.  Prove  that  when  Abelian  equations  are  irreducible,  they  are 
normal. 

Ex.  4.    Show  that  xn  —  1  =  0  is  Abelian  where  n  is  any  positive  integer. 

Ex.  5.  The  equation  x5  +  22  z4  -  440  a;3  -  3520  x  +  11264  x  +  32768  =  0 
has  as  three  of  its  roots  —  2,  4,  —  8.  Show  that  it  is  an  Abelian  equation. 


Ex.  6.  Is  z6  —  5  =  0  an  Abelian  equation  in  the  domain  fyi)  ?  In  the 
domain  0(1,  „,),  where  w  is  a  primitive  sixth  root  of  unity  ? 

185.  Abelian  Groups.  A  group  whose  substitutions  obey  the 
commutative  law  in  multiplication  is  called  an  Abelian  group. 
For  instance,  1,  (a&)  is  such  a  group,  because  1  •  (aft)  =  (a&)  •  1. 

210 


ABELIAN   EQUATIONS  211 

Ex.  1.    Every  sub-group  of  an  Abelian  group  is  itself  an  Abelian  group. 

Ex.  2.  If  GI  is  not  Abelian,  and  GI  is  a  sub-group  of  G,  then  G  is  not 
Abelian. 

Ex.  3.  Show  that  £3(3),  #2(4),  #4(4)  I,  #4(4)  H,  £4(4>  III,  G5<5),  G6&  II, 
are  Abelian  groups. 

186.  Abelian  Equations  have  Abelian  Groups.  If  the  roots  of 
an  Abelian  equation  are  all  distinct,  its  Galois  group  is  an 
Abelian  group. 

Let  f(x)  =  0  be  an  Abelian  equation,  and  let  its  roots  be 


If  f(x)  =  0  is  reducible,  let  g(x)  be  an  irreducible  factor,  and 
let  g(x)  =  0  have  the  roots 

a,  «'  =  <£'(«),  «"  =  $"(«),  -.  II 

All  the  roots  of  II  occur,  of  course,  in  the  series  I.  Now 
g(x)  =  0  satisfies  all  the  conditions  of  a  Galois  resolvent  of 
f(x)  =  0,  §  145.  Hence  the  group  of  f(x)  =  0  consists  of  the 
substitutions  p  _  (aa)>  pl  =  (aa,j}  ... 

This  group  obeys  the  commutative  law  in  multiplication,  for 


and,  §148,  P'P"  =  \a,  *' 
p"/={«,  <#," 
Since  the  equation  /(a;)  =  0  is  Abelian,  we  have 

*V(«)  «*'*"(«); 

hence,  P'P"  =  P"P'' 

Consequently,  the  group  of  substitutions  of  the  domain  fi(a) 
is  commutative,  as  is  also  the  isomorphic  group  of  the  equation 
/(a?)  =  0,  §  151.  Therefore,  the  Galois  group  of  f(x)  —  0  is  an 
Abelian  group. 


212  THEORY   OF   EQUATIONS 

187.    An  Equation  having  an  Abelian  Group  is  Abelian.      An 

irreducible  equation  g(x)  =  0,  having  a  commutative  group  is  an 
Abelian  equation. 

Let  a,  «!,  •••,  «n_i  be  the  roots  of  g(x)  =  0  and  let  G  represent 
the  group  of  this  equation.  As  g(x)  =  0  is  irreducible,  G  is 
transitive,  §  156. 

Let  s  be  any  substitution  in  the  group  G  which  does  not 
change  the  digit  0,  and  let  st  be  any  substitution  in  G  which 
replaces  0  by  i.  Then  s,"1  •  s  •  ss  is  a  substitution  of  G  which 
does  not  change  i ;  for 

sf~l  changes  i  to  0, 
s  does  not  change  0, 
st  changes  0  to  i. 

Since  the  group  G  is  assumed  to  be  commutative,  we  have 
s,.-1  •  s  •  Si  =  s,"1  •  s^  s  =  s. 

Hence  s  leaves  unchanged  not  only  the  digit  0,  but  also  the 
digit  i.  But  the  group  G  is  transitive ;  therefore,  the  digit  0 
must  be  capable  of  being  replaced  by  each  of  the  other  digits 
1,  2,  3,  •••,  (ft  — 1).  Yet,  no  matter  which  one  of  these  digits 
is  taken  to  be  i,  the  substitution  s  leaves  i  unaltered.  These 
relations  can  hold  true  only  when  s  is  the  identical  substitution 
in  the  group  G.  Hence  every  substitution  in  G,  except  1,  re- 
places 0  by  some  other  digit. 

Applying  to  every  other  digit  the  same  reasoning  which  we 
applied  to  0,  it  follows  that  every  substitution  in  the  group  G, 
except  the  substitution  1,  contains  that  digit  among  its  elements ; 
in  other  words,  there  is  no  substitution  in  G,  except  1,  which 
leaves  any  digit  unaltered. 

Next,  adjoin  to  the  domain  1}  the  quantity  M  =  a,  where  a 
is  one  of  the  roots  of  g(x)  —  0.  Since  no  substitution  in  the 
group  G,  except  1,  leaves  the  index  of  ax  unaltered  and  since 
the  identical  substitution  satisfies  the  definition  of  a  group,  1  is 
the  sub-group  to  which  M  belongs.  Thus,  Q  =  1 ;  and,  by  the 


ABELIAN   EQUATIONS  213 

adjunction  of  ax,  the  group  of  the  Galois  domain  is  reduced 
to  1,  §  163. 

The  Galois  domain  of  g(x)  =  0  is  fl(<v  ...  an_l},  §  143.  Each 
of  the  roots  «o,  a1}  •••,  «n_i  is  a  number  in  the  Galois  domain  and 
each  of  the  roots  admits  of  the  substitutions  of  the  sub-group 
Q  =  l;  hence  each  root  is  contained  in  the  domain  fi(a),  §  162, 
and  each  root  can  be  expressed  as  a  function  in  13  of  one  of 
them.  Therefore,  g(x)  =  0  is  a  normal  equation  and  the  domain 
O(a)  is  a  normal  domain,  §  132.  We  have  then 

«*  =  <£*(«), 
and  the  Galois  group  of  g(x)  =  0  consists  of  the  substitutions, 

§149'  /»*=(«,&(«)). 

We  have,  §  148,  Phpk  =  («,  &&(«)), 

p*ph  =  («,  <£*<£ft(«))- 
As  the  group  is  assumed  to  be  commutative,  we  must  have, 

<£*<£*(«)  =  &•£»(«)» 
i.e.  </(o;)  =  0  is  an  Abelian  equation. 

188.  Theorem.  In  a  substitution  belonging  to  a  transitive 
Abelian  group  all  the  cycles  consist  of  the  same  number  of 
elements. 

Let  the  substitution  s  be  resolved  into  its  cycles,  and  let  r 
be  the  least  number  of  elements  in  any  cycle.  The  substitution 
sr,  applied  to  the  elements  in  that  cycle,  leaves  the  elements 
unchanged.  Since,  §  187,  in  a  transitive  Abelian  group  no  sub- 
stitution, except  the  identical  one,  leaves  an  element  unaltered, 
sr  must  be  the  identical  substitution.  But  this  can  only  be  the 
case  when  all  other  cycles  (if  there  are  others)  consist  of  r 
elements. 

Ex.  1.  Name  the  Abelian  group  of  degree  five,  in  which  the  cycles  in 
one  and  the  same  substitution  do  not  have  the  same  number  of  elements. 
Explain.  See  Ex.  3,  §  185,  also  §  104. 


214  THEORY   OF   EQUATIONS 

Ex.  2.  Show  by  §§  187,  188  that  there  can  be  no  transitive  Abelian 
group  of  prime  degree  other  than  the  cyclic  group,  and  that  there  is  no 
irreducible  Abelian  equation  of  prime  degree  other  than  the  cyclic 
equation. 

Ex.  3.  Show  that  no  transitive  Abelian  group  of  degree  n  can  be  of 
lower  order  than  n. 

Ex.  4.  Show  that  a  transitive  Abelian  group  of  degree  n  is  of  the 
order  n.  Weber,  Vol.  I,  p.  578. 

189.  Solution  of  Abelian  Equations.  The  solution  of  Abelian 
equations  may  be  reduced  to  the  solution  of  cyclic  equations. 

In  a  transitive  Abelian  group  every  substitution,  except  the 
identical  one,  involves  all  the  elements  and  has  the  same  num- 
ber of  elements  in  each  cycle.  Hence,  if  n  is  the  total  number 
of  elements  and  r  is  the  number  in  one  cycle,  we  must  have 
n  =  r  •  t,  where  t  is  the  number  of  cycles  in  the  substitution. 

Let  O  be  the  group  of  an  irreducible  Abelian  equation  f(x)  —  0, 
and  let  s  be  any  substitution  except  1.  If  c,  cl}  «••,  c,^  are  the 
cycles  in  s,  we  may  write 

S  — •  CCjCg  *  *  *  Cj _j» 

Each  of  these  cycles  has  for  its  elements  r  roots  of  the  equa- 
tion f(x)  =  0.     Hence  we  have 

c  = 


where  the  a's,  /3's,  •••,  o-'s  are  the  roots  of  f(x)  =  0. 

Let  s1  be  any  substitution  in  the  group  G.     We  have,  §  187, 

S\       *  S  *  oj  — -  o» 

The  product  s1~lss1  is  obtained  by  performing  upon  each  cycle 
of  s  the  substitution  s1}  §  88.     As  this  operation  leaves  s  as  a 


ABELIAN   EQUATIONS  216 

whole  unchanged,  it  follows  that,  after  the  operation,  each 
cycle  still  has  the  same  letters  occurring  in  it  and  in  the 
same  cyclic  order,  though  the  cycles  may  have  interchanged 
positions.  Since  s  may  be  any  substitution  in  the  group  G, 
except  1,  we  conclude  that  the  group  is  imprimitive,  whenever 
t  >  1,  §  103. 

Let  M  be  a  cyclic  function  of  the  roots  a,  a1}  •  ••,  ar_l}  Ml  a 
cyclic  function  of  the  roots  fa  fa,  ...,  fa_lf  and  so  on.  We  have 
then 

M  =  !>«    «     •"    a_ 


There  will  be  t  such  conjugate  cyclic  functions,  M,  M1} 
M*  ••;  Mt_i. 

Let  Q  represent  the  aggregate  of  all  the  substitutions  in  the 
group  G  which  do  not  replace  a  cycle  by  another,  but  simply 
interchange  the  elements  in  each  cycle.  This  aggregate  of 
substitutions  is  a  group ;  the  product  of  any  two  of  them  gives 
a  substitution  belonging  to  G,  which  does  not  interchange  the 
cycles.  Thus,  Q  is  a  sub-group  of  G. 

As  no  substitution  in  Q  can  change  ak  into  any  element  not 
belonging  to  the  cycle  c,  Q  is  an  intransitive  group. 

The  function  M  is  readily  seen  to  admit  the  substitutions 
in  Q  and  those  only ;  hence,  if  we  adjoin  M  to  the  domain  Q, 
the  group  of  f(x)  =  0  reduces  to  Q,  §  163. 

As  Q  is  intransitive,  the  equation /(a?)  =  0  is  reducible  in  the 
domain  O(JO,  §  156. 

Let  f(x,  M)  be  a  function  of  x,  defined  thus : 

f(x,  M}  =  (x-a)(x—al}—(x-  «,._,). 

We  proceed  to  show  that  this  is  one  of  the  factors  of  f(x)  in 
the  domain  fi(Jf).  Since  Q  is  intransitive  and  permutes  the 
roots  in  each  cycle  among  themselves  only,  the  coefficients  of 
f(x,  M}  admit  all  the  substitutions  of  Q.  Therefore  f(x,  M) 


216  THEORY   OF   EQUATIONS 

is  a  function  of  x  in  O(JO,   §  154.      Since   all    the   roots   of 
f(x,  M")  =  0  are  roots  of  f(x)  =  0,  f(x,  M)  is  a  factor  of  /(*) 


in  O(J0. 

Similarly,  we  can  show  that 

f(x,  JMi)  =  (x-p)(x-K)...(x-  &_,), 
/(*,  Jf2)  =  (a?  -  y)  (»  -  Yl)  —  (a?  -  y^),  etc., 

are  factors  of  f(x).     We  have,  therefore, 


Since  the  coefficients  of  /(a;,  J!f)  =  0  .are  cyclic  functions  of 
its  roots,  the  group  of  this  equation  is  the  cyclic  group,  or  one 
of  its  sub-groups,  §  159.  But  a  cyclic  group  can  have  no  tran- 
sitive sub-group,  hence  the  irreducible  equation  f(x,  M)  =  0  is 
a  cyclic  equation.  Similarly  for  f(x,  J/i)  =  0,  etc. 

It  remains  to  explain  how  the  values  of  M,  •••,  Mt_±  may  be 
obtained.  By  §  161  they  are  roots  of  an  irreducible  equation 
g(M  )  =  0  in  O  of  the  degree  t.  We  proceed  to  prove  that 
g(M)  =  0  is  Abelian.  Since  f(x,  M}  =  0  is  cyclic,  we  get  for 
the  conjugates  of  Jf, 

M=t[a,  <f>(a),  .»,  ^-'(a)]  =F(a) 


By  assumption,  we  have  ^3  =  4>(«),  y  =  ^(cc).     Hence 


where  ^  admits  the  substitutions  of  the  cyclic  group.     Hence, 
by  §  162,  M!  is  a  function  in  ft  of  M.     Similarly  for  Mt. 


ABELIAN    EQUATIONS  217 

From  I  we  see  that  replacing  «  by  ft  or  y  changes  M  into  Ml 
or  M2.     Hence,  if 


Ml  =  \(M)  =  F$  (a),  M2  =  ^(M)  =  F^>1  (a), 
we  may  write 

=  \A!  (M)  = 


Since,  by  assumption,  ^(a)  =  $yl>(a),  we  have  also 

=  A1X(Jf).     Similarly  for  other  conjugates  of  M.     We  have 

now  proved  that  g(M}  =  0  is  an  Abelian  equation. 

Hence  we  have  shown  that  the  solution  of  the  given  Abelian 
equation  /(a?)  =  0  can  be  reduced  to  the  solution  of  cyclic  equa- 
tions and  of  another  Abelian  equation  of  lower  degree.  The 
latter  Abelian  equation  -can  be  treated  in  the  same  manner 
as  was  f(x)  ==  0  ;  hence,  eventually,  the  solution  of  f(x)  =  0  is 
reduced  to  that  of  cyclic  equations  only. 

Ex.  1.    Abel  gave  the  following  example  of  an  Abelian  equation.    Let 

2  7T 

a  =  —  ;  then  cos  a,  cos  2  a,  •••,  cos  na  can  be  shown  to  be  the  roots  of 
n 

the  equation       ^  _  n  xn_2  +  J_  n(n  -  3)  xn-4  +  ..  .  _  0.  I 

For  the  derivation  of  this  equation  see  Serret's  Algebra  (Ed.  G. 
Wertheim),  1878,  Vol.  I,  pp.  195-199.  Expanding  the  right  member 
of  De  Moivre's  formula,  cos  ma  +  i  sin  ma  =  (cosa  +  t'sin  a)m,  by  the 
binomial  theorem,  we  can  express  cos  ma  as  a  function  in  ii(i)  of  cos  a. 
We  may,  therefore,  write  cos  mot  =  0(cos  a),  where  6  is  the  function. 
Similarly,  cos  mta  =  ^(cos  a).  Writing  mi«  for  a  in  the  former  equation, 

-rrrp    Dfftt 

=  0(cos  mta)  =  00!  (cos  a). 


If  in  01(cos  a)  =  cos  m^a  we  replace  a  by  ma,  we  have 
cos  (mjTOrt)  =  0!(cos  ma)  =  0^(003  a). 

Hence  every  root  of  I  can  be  expressed  as  a  function  in  Q  of  one  of 
them,  and  we  have  in  addition 

00i(cos  a)  =  0!0(cos  a). 
Therefore  I  is  an  Abelian  equation. 


218  THEORY   OF    EQUATIONS 

Ex.  2.  Show  that  I  in  Ex.  1  is  a  reducible  equation  in  the  domain  0 
defined  by  its  coefficients. 

Consider  the  value  of  the  root  cos  na. 

Ex.  3.  The  equation  z*  + 1  =  0  has  the  group  P  =  <?4(4>II,  §  159,  Ex.  6. 
Its  roots  are  a  =  J  V2(l  +  i),  «i  =  -  Jv^l  -  i) ,  «2  =  -  a,  «3  =  -  at. 
Illustrate  the  reduction  of  the  solution  of  Abelian  equations  to  that  of 
cyclic  equations. 

Lets=  (««1)(«2«3),  «  =  (««!))  ^  =  (0203),  -5f=«a12+«1«2,  2tf1=a3a22 
4-  «2«32j  §  =  1)  (««i)  («2«s).  Here  J!f  and  MI  are  the  roots  of  t2  +  2  =  0 ; 
i.e.  M  =  i  V2,  Ml=-  i  V§.  Then  /(»,  i)  =  x2  +  i  =  0,  /(x,  -  i)  =  x2 
—  i  =  0  are  both  cyclic  equations. 

Ex.  4.  The  equation  3*  -  8  x8  +  20  x2  - 16  x  + 1  =  0  has  the  Galois  group 
G^WII ;  hence,  is  irreducible  and  Abelian.  We  have  here  «j  =  —  a  +  4, 
02  =  _  a8  +  6  a2  -  8  a  +  2,  «3  =  a3  -  6  a2  +  8  «  +  2.  Illustrate  the  re- 
duction, as  in  Ex.  1.  Netto,  Algebra,  Vol.  II,  p.  234. 


CHAPTER  XIX 

THE  ALGEBRAIC   SOLUTION   OF  EQUATIONS 

190.  Adjunction  of  Roots  of  Binomial  Equations.  In  this 
chapter  it  is  proposed  to  develop  the  necessary  and  sufficient 
conditions  for  the  solvability  of  algebraic  equations  of  any 
degree.  To  this  end  we  shall  assume  in  this  paragraph  that 
f(x)=Q  is  an  equation  which  admits  of  being  solved  by  algebra; 
that  is,  we  shall  assume  that  all  the  roots  of  the  given  equation 
f(x)  =  0  can  be  obtained  from  its  coefficients  by  a  finite  number 
of  additions,  subtractions,  multiplications,  divisions,  and  ex- 
tractions of  roots  of  any  index. 

Let  Vc,  where  c  is  an  algebraic  number,  be  any  one  of  the 
radicals  which  enter  into  the  expressions  for  the  roots  of 

G2 
a,  «!,  •••,  «„_!  of  the  equation  f(x)=0.     Thus,  if  c  =  —  +H3 

and  m  =  2,  then  Vc  is  one  of  the  radicals  appearing  in  the 

G 

solution  of  the  cubic,  §59.     Ifc  =  —  — 

2i 

have  another  radical  entering  the  expression  of  the  roots  of  a 
cubic.  Now  the  rath  power  of  any  radical  Vc  is  a  number  in 
the  domain  O(c).  In  other  words,  every  radical  is  a  root  of  a 
binomial  equation  of  the  form  of  —  a  =  0.  Thus  it  is  evident 
that  all  the  radicals  which  go  to  make  up  a  root  of  f(x)  =  0  are 
roots  of  binomial  equations. 

If /(a)  =  0  is  reducible  in  the  domain  to,  defined  by  its  coef- 
ficients, we  may  apply  to  its  irreducible  factors  the  argument 
which  follows.  If  f(x)  =  0  is  irreducible  in  that  domain,  it  is 

219 


220  THEORY   OF    EQUATIONS 

clear  that  by  the  successive  adjunction  of  some  or  all  the  radi 
cals  which  enter  into  the  expressions  for  its  roots,  the  equation 
will  become  reducible  in  the  enlarged  domain.  That  is,  f(x)=0 
becomes  reducible  upon  the  adjunction  of  certain  roots  of  binomial 
equations. 

As  an  illustration,  observe  that  in  §  167  the  solution  of  the 
quadratic  equation  was  made  to  depend  upon  the  adjunction 
of  y,  the  root  of  the  binomial  equation  y2=a12—  4  0%. 

In  the  case  of  the  cubic,  §  168,  we  first  adjoined  V-D,  which 
is  the  root  of  a  binomial  equation  obtained  by  removing  the 
second  term  from  the  quadratic  ue  -\-  Gu3  —  H3  =  0.  Next  we 
adjoined  u,  which  is  a  cube  root  of  a  binomial. 

In  the  case  of  the  quartic,  §  169,  we  first  adjoined  u,  which 
differs  only  by  a  rational  constant  from  a^.  Here  xl  is  the 
root  of  a  cubic  equation,  the  solution  of  which  may  itself  be 
explained  by  the  adjunction  of  roots  of  binomial  equations,  as 
we  have  just  seen.  Next  we  adjoined  Vi>,  Vw,  Vtu,  all  roots 
of  binomial  equations. 

191.  Dependence  upon  Cyclic  Equations.  All  binomial  equa- 
tions are  known  to  be  Abelian  equations,  §  184,  Exs.  4,  6,  and 
Abelian  equations  may  be  solved  algebraically  by  the  aid  of  a 
series  of  cyclic  equations  whose  degrees  are  prime,  §  189. 
Consequently,  when  f(x)  =  0  is  a  solvable  equation,  its  solution 
may  be  made  to  depend  upon  that  of  cyclic  equations  of  prime 
degree. 


192.  Restatement  of  the  Problem.  Suppose  now 
is  any  algebraic  equation.  The  question,  whether  it  is  solvable 
by  radicals,  may  be  replaced  by  the  question  of  equal  scope, 
whether  it  is  solvable  by  roots  of  cyclic  equations  of  prime 
degree.  We  have  thus  arrived  at  the  following  query  :  Under 
what  conditions  is  the  group  O  of  an  equation  of  the  nth  degree, 
f(x)  =  0,  reduced  by  the  adjunction  of  a  root  of  a  cyclic  equation 
whose  degree  is  prime  ¥ 


THE   ALGEBRAIC    SOLUTION    OF    EQUATIONS          221 

193.  Theorem.  If  the  group  G  of  an  equation  f(x)  =  0  is 
reduced  by  the  adjunction  of  a  root  of  a  cyclic  equation  of  the 
prime  degree  m,  then  the  group  G  has  a  normal  sub-group  whose 
index  is  the  prime  number  m. 

Let  f(x)  =  0  be  reducible  or  irreducible,  but  free  of  multiple 
roots.  Let  h(x)  =  0  be  a  cyclic  equation  of  the  mth  degree, 
where  m  is  a  prime  number.  We  assume  that  the  adjunction 
of  one  of  the  roots  of  h(x)  =  0  does  reduce  the  group  G  to  one 
of  its  sub-groups  Q. 

Let  the  roots  of  h(x)  =  0  be  X,  Xlt  -••,  Xm_^  Since  h(x)  =  0 
is  cyclic,  all  its  roots  can  be  expressed  as  functions  in  fl  of  one 
of  them.  If  G  is  the  group  of  f(x)  =  0  in  O,  then  Q  is  the 
group  of  the  same  equation  in  the  domain  fi(X),  or  in  the 
coextensive  domains  O(x),  •••,  13(Xn,_,)- 

According  to  §  165,  Cor.  II,  the  degree  m  of  h(x)  =  0  is  a 
multiple  of  j,  the  index  of  the  group  Q  under  G.  Since  m  is  a 
prime  number,  and  j  must  be  greater  than  1,  we  have  m  =  j. 

Let  M  be  a  function  in  O  of  the  roots  of  f(x)  =  0,  and  let 
M  belong  to  the  sub-group  Q.  Then  M  is  a  function  in  fi  of  X, 
§165. 

Again,  by  §  165,  Cor.  I,  the  domain  of  O(JO  is  a  divisor  of  the 
domain  O(x).  But  the  degree  of  OU)  is  prime,  being  by  defini- 
tion, §  132,  of  the  same  degree  as  that  of  the  equation  h(x)  =  Q, 
which  has  X  as  a  root. 

Since  fi(JO  is  a  divisor  of  Ow,  and  the  degree  of  O(J)  is  prime, 
we  must  have  O(X)  =  1)(JO.  Hence,  not  only  is  M  a  function  in 
O  of  X,  but  X  is  a  function  in  O  of  M,  and  either  function 
admits  of  all  the  substitutions  that  the  other  does.  Hence  X, 
like  M,  belongs  to  the  group  Q. 

Operate  upon  X  with  the  substitutions  of  G,  and  we  get  the 
following  distinct  values :  X,  X\,  •••,  X^.  By  §  161  these 
values  are  roots  of  an  irreducible  equation.  This  must  be 
identical  with  the  irreducible  equation  h(x)  =  0,  since  the  two 
have  the  root  X  in  common,  §  126.  Thus,  the  values  X,  Xlt 
>",  Xm.lf  and  X,  X'^  •••,  X'w_t,  are  equal  respectively. 


222  THEORY   OF    EQUATIONS 

Let  s  be  a  substitution  in  G  which  changes  X  to  Xt.  That 
same  substitution  transforms  the  sub-group  Q  into  the  conjugate 
sub-group  s^Qs^Q^  Now  the  substitutions  in  the  sub-group 
Q,  leave  Xi  unchanged.  For,  to  operate  with  the  substitutions 
in  Q!  is  the  same  as  to  operate  with  s~lQs,  where  s~l  changes 
Xi  to  X,  and  X  remains  unaltered  by  the  substitutions  in  Q, 
while  s  changes  X  back  to  X^  But  X  and  Xl  are  roots  of  a 
cyclic  equation ;  hence  Xt  is  a  function  in  Q  of  X,  and  X  is  a 
function  in  fl  of  X,  so  that  X  and  X!  belong  to  the  same  group 
Q.  Therefore,  Q  =  Qi. 

Since  the  same  reasoning  applies  to  X  and  any  one  of  the 
other  roots  X2,  •••,  Xm_l}  it  follows  that  Q  is  identical  with  all 
of  its  conjugate  groups ;  that  is,  Q  is  a  normal  sub-group  of  G, 
having  the  index  m. 

194.  The  Converse  Theorem.  If  the  group  G  of  the  equation 
f(x)  =  0  has  a  normal  sub-group  Q,  ivhose  index  is  a  prime  num- 
ber m,  then,  by  adjunction  of  a  root  of  a  cyclic  equation  of  the 
degree  m,  the  group  G  is  reduced  to  Q. 

If  the  group  G  has  a  normal  sub-group  Q  of  the  prime  index 
m,  and  if  we  select  a  function  M  which  belongs  to  the  sub- 
group Q,  the  conjugate  functions  all  belong  to  the  same  group 
Q.  By  §  162,  each  function  M,  Mlt  •••,  Mm_i,  is  contained  in 
the  domain  13(jr).  Hence  this  domain  is  a  normal  domain,  §  132, 
and  M  is  the  root  of  a  normal  equation,  §  139.  In  the  domain 
Q(JO  we  have  Q  as  the  group  of  the  equation  f(x)  =  0,  §  163. 
But,  if  m  is  a  prime  number,  the  normal  equation  is  also  a 
cyclic  equation;  for,  the  degree  m  of  the  normal  equation  is 
also  the  order  of  the  Galois  group,  §§149,  150.  Take  any 
substitution  s  (not  the  identical  substitution)  in  the  Galois 
group.  The  different  powers  of  s  constitute  a  sub-group,  the 
order  of  which  is  a  factor  of  the  order  of  the  Galois  group. 
As  m  is  prime,  the  order  of  s  must  be  m  and  the  sub-group  is 
s,  s2,  s3,  •••,  sm.  The  Galois  group  and  its  sub-group,  being  of 


THE   ALGEBRAIC   SOLUTION   OF   EQUATIONS          223 

the  same  order,  are  identical.  Hence  the  Galois  group  is  the 
cyclic  group,  s,  s2,  •••,  sm,  and  the  normal  equation  is  a  cyclic 
equation,  §  170. 

195.  Metacyclic  Equations.     An  equation  is  called  metacyclic 
or  solvable,  when  its  solution  can  be  reduced  to  the  solution  of 
a  series  of  cyclic  equations.     Abelian  equations  are  a  special 
class  of  metacyclic  equations.    The  latter  embrace  all  equations 
that  are  solvable  by  radicals,  and  no  others. 

In  §  191  it  was  shown  that  any  equation  which  can  be  solved 
by  radicals  can  be  solved  by  the  aid  of  cyclic  equations  of 
prime  index.  In  §  193  it  was  shown  that  if  the  adjunction  of 
a  root  of  a  cyclic  equation  of  prime  degree  reduces  the  group 
G,  there  exists  a  normal  sub-group  whose  index  is  a  prime 
number ;  while  in  §  194  it  was  shown  that,  if  G  has  a  normal 
sub-group,  the  reduction  can  always  be  effected  by  the  adjunc- 
tion of  such  a  root. 

196.  Criterion  of  Solvability.     TJiat  a  given  algebraic  equation 
be  metacyclic  it  is  necessary  and  sufficient  that  there  exist  a  series 
of  groups  G,  G»  G2, -,  Gk  =  l, 

the  first  of  which  is  the  Galois  group  of  the  equation  in  O,  the  last 
of  which  is  the  identical  group,  each  group  being  a  normal  sub- 
group of  the  preceding  and  of  a  prime  index. 

The  group  G  of  a  metacyclic  equation  must  have  a  normal 
sub-group  of  an  index  j  that  is  a  prime  number.  Call  this  sub- 
group GI.  If  6?!  consists  of  the  identical  substitution  only 

(whose  order  is  1),  then  j  ==  j  •     That  is,  the  order  of  G  itself 

is  prime,  and  G  has  no  sub-groups,  except  1.  This  can  happen 
only  when  G  itself  is  a  cyclic,  group,  and  the  given  metacyclic 
equation  is  itself  only  a  cyclic  equation. 

If  GI  is  not  1,  then,  since  the  equation  is,  by  hypothesis, 
solvable  by  radicals,  Gl  must  again  have  a  normal  sub-group  G^ 


224  THEORY  OF   EQUATIONS 

whose  index  is  a  prime  number  j2.  Continuing  in  this  way,  we 
finally  arrive  at  the  identical  group  1.  This  proves  the 
theorem. 

197.  Criterion  Applied.     The  Galois   group   of  the   general 
equation  of  the  nth  degree  is  the  symmetrical  group  of  the  nth 
degree.     The   symmetric   group   has   always   the   alternating 
group  as  a  sub-group.     This  alternating  sub-group  is  a  normal 
sub-group  of  the  index  2.     It  becomes  the  group  of  the  given 
equation  by  the  adjunction  of  the  square  root  of  the  discrimi- 
nant.   The  principal  series  of  composition,  §  110,  is  G2(<2),  1,  for 
the  quadratic;  6r6(3),  Gr3(3),  1,  for  the  general  cubic;  and  Gr24(4), 
C?i2(4),  (?4(4)  II,  6r2(4),  1,  for  the  general  quartic.    In  these  cases  the 
alternating  group  is  seen  to  have  a  normal  sub-group  of  prime 
index.     We  are  going  to  show  that  when  the  degree  of  the 
general   equation   is   greater   than  4,  and,  consequently,  the 
degree  of  the  Galois  group  is  greater  than  4,  the  alternating 
group  has  no  normal  sub-group  of  prime  index. 

198.  Theorem.     An  alternating  group  of  higher  degree  than 
the  fourth  has  no  normal  sub-group  of  prime  index. 

All  substitutions  of  an  alternating  group  are  even,  §§  99, 100, 
and  are  expressible  as  the  product  of  cycles  of  three  elements, 
§  93.  Let  these  substitutions  be  so  expressed. 

We  first  establish  the  possibility  of  selecting  a  substitution 
s  in  the  alternating  group,  so  that  a  given  cycle  of  three 
elements,  say  (1  2  3),  will  be  transformed  into  any  other  cycle 
of  three  elements  occurring  in  the  alternating  group.  Suppose 
that  1,  2,  3,  4,  r,  t,  u,  v,  are  elements  of  the  group  and  we  wish 
to  transform  (1  2  3)  into  (r  t  u).  It  is  easily  seen  that  the 
substitution  *=  A  234\  will  do  it;  for,  8-l(12S)s  =  (rtu). 

That  s  is  a  substitution  in  the  alternating  group  is  clear,  since, 
§  82,  s  =  (1 2 1)  (1 2  r)  (3  4  v)  (3  4  w),  an  even  substitution. 


THE   ALGEBRAIC   SOLUTION    OF   EQUATIONS          225 

Next,  let  Q  be  a  normal  sub-group  of  the  alternating  group, 
let  Si  .be  any  substitution  in  Q  (except  the  substitution  1),  and 
s  any  substitution  in  the  alternating  group.  It  is  easy  to  see 
that,  by  the  property  of  normal  sub-groups,  s'^s  is  also  a 
substitution  in  Q. 

If  Si  consists  of  a  cycle  of  three  elements,  we  can,  by  proper 
selection  of  s  in  the  operation  s~lsls,  transform  s^  into  any 
other  cycle  of  three  elements.  Therefore,  Q  must  contain  all 
cyclic  substitutions  of  three  elements  whenever  it  contains  one 
of  them,  and  must,  consequently,  be  identical  with  the  alter- 
nating group. 

Since  s^1  and  s'^s  are  both  substitutions  in  Q,  their  product 
must  be ;  namely,  ^  _  s  -i .  s-is  s> 

We  shall  now  show  that,  whenever  n  >  4,  s  can  always  be 
chosen  from  the  substitutions  of  the  alternating  group  in  such 
a  way  that  the  substitution  A.  represents  a  cycle  of  three  ele- 
ments, thereby  showing  that  the  normal  sub-group  Q  is  really 
identical  with  the  alternating  group ;  in  other  "words,  showing 
that  there  is  no  normal  sub-group,  distinct  from  the  alternating 
group  itself,  except  the  group  1. 

To  show  this,  we  assume  that  all  the  substitutions  in  the 
alternating  group  and  in  Q  are  resolved  (as  they  always  can 
be)  into  cycles  so  that  no  two  cycles  have  an  element  in  com- 
mon, §  86.  In  the  formation  of  X  there  is  no  need  whatever 
of  considering  those  cycles  in  the  substitutions  Sj  whose  ele- 
ments are  unaffected  by  s,  because  in  the  product  s^'s"1^  they 
cancel  each  other.  We  shall  consider  separately  the  different 
forms  which  sr  may  take,  when  n  >  4. 

(1)  Let  some  one  substitution  st  in  the  normal  sub-group  Q 
have  a  cycle  (123---m)  which  consists  of  more  than  three 
elements.  Then  s1=(l  2  3  •••  ra)^  •••,  where  cl5  c2,  •••  are  cycles 
which  do  not  contain  the  elements  1  2  3  •  •  •  m.  Choose  s = (1 2  3), 
then  srls-\  =  Sl-\l  3  2)  sl  =  (2  4  3),  and  \=sl-ls-lsls=(2  4  3). 
(1  2  3)  =  (1  2  4).  Hence  Q  contains  a  substitution  A.  consisting 
Q 


226  THEORY  OF   EQUATIONS 

of  a  cycle  of  three  elements,  and  therefore  Q  is  identical  with 
the  alternating  group.  Thus,  there  is  no  normal  sub-group 
containing  the  substitution  (123  •••  m)c1c2'--. 

(2)  Let  some  one  substitution  s^^  in  Q  consist  of  two  or  more 
cycles,  two  cycles  of  which  contain  each  three  elements.     Let 
these  two  cycles  be   (1  23)  (4  5  6).      Take  s  =  (1  3  4),   then 
ar1*-1*!  =  (2  5  1)  and  A.  =  (2  5  1)  (1  3  4)  =  (1  2  5  3  4).     This  sub- 
stitution A.,  found  in  Q,  has  more  than  three  elements  in  its 
cycle,  and  comes  under  case  (1).     Hence,  there  is  no  normal 
sub-group  of  the  alternating  group  containing  a  substitution 
Sl  =  (1  2  3)  (4  5  6). 

(3)  Let  s1  consist  of  cycles,  embracing  one  cycle  of  three  ele- 
ments and  another  of  two  elements,  viz.,  the  cycles  (1  2  3)  (4  5). 
Choose  s  =  (1  2  4),  then  A  =  (2  5  3)  (1  2  4)  =  (1  2  5  3  4),  which 
comes  under  case  (1).     Hence,  there  is  no  normal  sub-group 
containing  st  =  (1  2  3)  (4  5). 

(4)  Let  Sj   embrace ,  three    transpositions,    (1  2)  (3  4)  (5  6). 
Choose  s  =  (1  3  5),  then  A.  =  (2  6  4)  (1  3  5),  which  comes  under 
case  (2).     Thus  the  possibility  of  the  existence  of  a  normal 
sub-group,  containing  s^  =  (1  2)  (3  4)  (5  6),  is  excluded. 

(5)  Let  Sj  consist,  in  part  or  wholly,  of  two  transpositions 
and  one  invariant  element.     That  is,  let  ^  contain  among  its 
cycles  (1 2) (3  4)(5).   Take  s=  (1  2  5)  and  we  get  X  =  (1  2  5)(1 2  5) 
=(15  2).     Hence,  Q  again  coincides  with  the  alternating  group. 

The  above  cases  exhaust  all  the  cases  which  are  possible 
when  n  >  4. 

When  w==4,  a  new  possibility  arises ;  namely,  sl=(H  2)  (3  4). 
No  matter  what  substitution  in  the  alternating  group  G12(4)  is 
chosen  for  s,  we  fail  to  get  for  A  a  cycle  of  three  elements.  On 
the  other  hand,  the  sub-group 

1,  (1  2)(3  4),  (1  3)(2  4),  (1  4)(2  3) 

satisfies  the  characteristic  property  of  a  normal  sub-group  of 
ft  (4) 


THE   ALGEBRAIC   SOLUTION   OF   EQUATIONS          227 

The  group  1  is  a  normal  sub-group  of  any  group,  but  it  is  not 
a  normal  sub-group  of  prime  index  for  alternating  groups  of 
degrees  higher  than  the  fourth.  The  order  of  the  alternating 

group  of  the  nth  degree  is  tS  •     Now  l^  -*- 1  is  the  index  of  the 

2  2 

group  1  under  the  alternating  group.  When  n  >  4,  this  index 
never  is  a  prime  number.  Hence  the  theorem  is  established. 

199.  Insolvability  of  General  Equations  of  the  Fifth  and  Higher 
Degrees.     From  §§  196,  198,  it  appears  that  the  general  equa- 
tions of  higher  degree  than  the  fourth  do  not  satisfy  the  con- 
ditions of  solvability.    However,  a  special  equation  of  a  higher 
degree  than  the  fourth,  whose  group  is  not  the  symmetric  or 
the  alternating  group,  may  possess  the  necessary  series  of  nor- 
mal sub-groups  of  prime  index,  and  may  be  solvable  by  radicals. 
Thus,  any  equation  of  the  fifth  degree  whose  group  is  not  the 
symmetric  or  alternating  group  can  be  solved  by  radicals. 

Of  the  295  substitution-groups  whose  degree  does  not  exceed 
eight,  only  28  are  insolvable.  See  Am.  Jour,  of  Math.,  Vol.  21, 
p.  326. 

Ex.  1.  Show  that  the  quartic  in  Ex.  9,  §  159,  is  metacyclic,  but  not 
Abelian  ;  find  its  principal  series  of  composition. 

200.  A  Criterion  of  Metacyclic  Equations  of  Prime  Degree.    All 

algebraic  equations  of  the  first  four  degrees  are  metacyclic. 
The  following  process  enables  one  to  ascertain  whether  a  given 
equation  of  the  fifth  or  a  higher  prime  degree  is  metacyclic  or 
not. 

If  the  given  irreducible  equation  f(x)  =  0  is  metacyclic,  then 
one  of  the  series  of  groups  O,  Gv  •••,  Gk~in  §  196  must  be  the 
Galois  group  of  the  given  equation.  Proceeding  as  in  §  159; 
let  «o,  «!,  •••,  «„_!  be  its  roots;  also  let  y  be  a  function  of  «o, 
«i>  "•)  ««-!»  formally  unaltered  by  the  substitutions  in  O  and 
those  only,  where  G  is  the  group  of  highest  order  in  this  series. 
Let  the  index  of  G  with  respect  to  the  symmetric  group  of 


228  THEORY    OF   EQUATIONS 

degree  n  be  j.  Operating  upou  y  with  the  substitutions  of  the 
symmetric  group  we  get  j  expressions  for  y,  distinct  in  form, 
viz.,  y^  y?,  •••>  %••  Construct  the  equation  of  degree./, 


The  coefficients  of  I  are  symmetric  functions  of  the  roots, 
«o>  •••,  «»_i5  hence  they  are  rational  in  O  and  can  be  computed. 

If  in  the  function  y  we  substitute  the  values  of  the  roots  of  a 
metacyclic  equation  of  the  nth  degree,  y  assumes  a  numerical 
value  which  lies  in  O.  For,  assuming  that  the  equation  is 
metacyclic,  its  Galois  group  must  be  either  G  or  one  of  the  sub- 
groups Gl}  -•',  GK  §  196;  hence  y  admits  the  substitutions  of 
the  Galois  group  and  is,  therefore,  a  number  in  O,  §  154. 

Conversely,  if  the  function  y  becomes  a  number  in  fl,  when 
the  values  of  the  roots  of  /(a?)  =  0,  n  being  prime,  are  substi- 
tuted in  it,  so  that  I  has  a  rational  root,  which  is  not  a  multiple 
root,  then  is  f(x)  =  0  metacyclic.  For,  under  these  condi- 
tions y  belongs  to  G,  and  the  Galois  group  of  f(x)  =  0  must 
be  either  G  or  one  of  its  sub-groups,  §  159.  If  it  is  G,  then  the 
conclusion  follows  at  once  ;  if  it  is  one  of  its  sub-groups,  it  can 
be  shown  (the  proof  is  here  omitted)  that,  when  n  is  prime,  the 
sub-group  is  one  of  the  metacyclic  groups  Gi,  G2,  •••,  Gk_i,  so 
that  f(x)  =  0  is  a  metac^clic  equation.* 

Hence  the  rule  :  Select  a  function  y,  formally  unaltered  by  the 
substitutions  in  G,  and  those  only,  so  that  F(y)  =  0  has  no  mul- 
tiple roots.  If  F  (y)  =  0  has  a  rational  root,  f(x)  =  0  is  meta- 
cyclic, otherwise  it  is  insolvable. 

Theoretically,  it  matters  not  what  function  of  OQ,  «T,  •••,  «„_! 
is  selected  for  y,  if  only  it  belongs  to  the  group  G.  Practically, 
much  depends  upon  this  selection,  as  the  algebraic  operations 
are  very  much  more  complicated  with  some  functions  than  with 
others.  The  computation  of  the  coefficients  of  F(?/)  =  0  is 

*  For  a  complete  discussion  see  H.  Weber,  Algebra,  Vol.  I,  1898,  §§  188,  or 
E.  Netto,  Algebra,  Vol.  H,  1900,  §  611-615. 


THE   ALGEBRAIC   SOLUTION   OF   EQUATIONS          229 

usually  very  laborious  even  in  the  case  of  the  quintic.  Inas- 
much as  Bring,  in  1786,  and  Jerrard,  in  1834,  were  able  to  trans- 
form the  general  quintic  to  the  form  a?  +  cx  +  d  =  Q  (for 
this  transformation,  see  Netto's  Algebra,  Vol.  I,  pp.  124,  125), 
it  is  of  interest  to  compute  F(y)  =  0  for  this  special  form. 

Ex.  1.  Find  the  condition  that  the  equation  y£  +  ex  +  d  =  0,  when 
irreducible,  shall  be  metacyclic. 

Referring  to  §  104,  we  see  that  for  the  quintic  the  metacyclic  group  of 
the  highest  order  G  is  (abcde)20.  As  a  function  belonging  to  this  group 
select  (following  C.  Runge,  Acta  Math.  1  (1885),  p.  173)  y2,  where 

y  =  a0«l  +  «1«2  +  «2«3  +  «3«4  +  «4«0  —  «0«2  ~  «2«4  ~  «4«1  —  «1«3  —  «3«0. 

Here  j  =  6  and  F(y)  =  0  is  a  resolvent  equation  of  the  sixth  degree. 
We  find  it  convenient  to  consider  y  itself,  which  is  not  a  metacyclic  func- 
tion. Operated  upon  by  the  symmetric  group,  y  yields  twelve  values,  of 
which  six  differ  from  the  other  six  simply  in  sign.  Let  one  set  of  six 
values  be  j/i,  z/2,  •••,  y6.  Also  let  the  equation  of  which  they  are  roots  be 

y6  +  ciiy5  +  «2y*  +  a3y*  +  a4y2  +  asy  +  a6  =  0.  I 

Its  coefficients  a\,  02,  •••,  «e  are  n°t-  necessarily  rational  numbers,  but 
they  are  symmetric  functions  of  j/i,  •••,  y6.  Consider  yi,  .-,  y6  as  func- 
tions of  «o,  •••,  «n-i,  and  operate  upon  them  with  the  alternating  group ; 
the  values  y^  •  ••,  y6  are  merely  permuted  among  themselves.  Substitu- 
tions which  do  not  belong  to  the  alternating  group  bring  about  a  change 
in  sign.  The  coefficients  a\,  a^  •••,  «e  are  therefore  either  symmetric  or 
alternating  functions  of  «o,  •••,  a»_i.  Of  these  a2,  «4,  «e  are  symmetric 
functions  because,  being  homogeneous  functions  of  even  degree,  they  are 
not  affected  by  changes  of  signs  in  yi,  yz,  •••,  y6.  On  the  other  hand, 
«i>  «3i  «5  are  alternating  functions  of  «o,  «i,  •••,  On-:,  being  homogeneous 
functions  of  odd  degree. 

If  D  is  the  discriminant  of  the  quintic,  then  Vl)  is  a  function  of 
Oo,  •••,  «n-i  belonging  to  the  alternating  group.  Hence  the  coefficients 
«i,  as,  05  are  of  the  form  miVZ),  m2VZ),  wisVZ),  where  mi,  m2,  ma  are 
symmetric  integral  functions.  With  respect  to  «o,  «it  •••)  #n-i,  it  is  seen 
that  «i  is  of  the  second  degree.  But  ai  is  also  of  the  form  wiiVZ),  where 
mi  is  integral  and  VD  is  of  the  tenth  degree.  Hence  we  must  have 
mi  =  0.  Similarly,  a3  being  of  the  sixth  degree,  yields  m3  =  0.  On  the 
other  hand,  a6  and  VZ)  are  both  of  the  tenth  degree.  Write  a$  =  mv/ZX 
Equation  I  becomes 


230  THEORY   OF   EQUATIONS 

In  the  equation  x5  +  ex  +  d  =  0,  c  and  d  are  homogeneous  functions  oi 
the  roots  of  the  degrees  4  and  5,  respectively.  Since  a2,  a4,  a$  are  of  the 
degrees  4,  8,  12,  we  may  write 

a2  =  mzc,  a4  =  rw4c2,  ae  =  wigc3, 

where  m2,  m4,  wie  are  integers.  To  find  the  values  of  m,  m2,  m4,  m6, 
assign  to  c  and  c?  the  special  values  c  =  —  1,  d  =  0.  Then  Z)  =  —  44  ;  the 
five  roots  are  0,  t,  t2,  i3,  i4  ;  the  six  values  yi,  •  ••,  y6  are  —  2  i,  —  2  i,  —  2  i, 
—  2  i  ,  2  +  4  i,  —  2  +  4  t.  Equation  II  becomes 

0  =  y6  -  m2y4  +  m4y2  -  m6  +  16  M»y  =  (y  +  2  i)4(y2  -  8  iy  -  20) 

=  y*  +  20  y4  +  240  z/2  -  320  +  512  iy. 

Hence  m2  =  —  20,  m4  =  240,  m6  =  —  320,  m  =  32.  Substituting  in  II, 
and  squaring  to  remove  the  radical,  we  have 


(y6  -  20  c?/*  +  240  c2y2  +  320  c3)2  =  45  Zty2,  III 

or  (y2  -  4  c)*(y4  -  24  cy2  +  400  c2)  =  4^  .  55  •  d4?/2, 

where  J5  =  44c5  +  55t?4.     Write  y2  =  4  ^  ;  then  y2  being  metacyclic,  so  is  2. 

We  obtain 

(z3  -  5  cz2  +  15  c2z  +  5  c8)2  =  Dz,  IV 

which  may  also  be  written 


If  x5  4-  ex  +  d  =  0  is  irreducible,  it  is  metacyclic  when  IV  or  V  has  a 
rational  root,  and  then  only.  If  the  quintic  is  reducible,  it  is  always 
solvable.  For  a  different  treatment  of  the  quintic  see  Glashan  and  Young 
in  Am.  Jour,  of  Math.  7  (1885),  and  especially  McClintock,  ib.  8  (1886) 
and  20  (1898). 


Ex.  2.  Show  that  no  equation  of  the  form  x6  +  5x  +  5t  =  Q  is  meta- 
cyclic, where  t  is  any  integer  not  a  multiple  of  5. 

By  §  129,  the  equation  is  irreducible.  If  IV  in  Ex.  1  has  in  this  instance 
a  rational  root,  it  must  be  integral,  since  the  coefficients  of  the  quintic  are 
integral  and  the  first  term  is  x5.  It  must  also  be  a  divisor  of  the  absolute 
term  25  c6  or  58.  But  no  factor  of  58  is  a  root  of  the  equation. 

Ex.  3.    Show  that  a^  +  15  x  +  12  =  0  is  irreducible  and  metacyclic. 

Ex.  4.  Is  x5  +  5  x4  +  10  x3  +  10  xz  +  7  x  +  5  =  0  metacyclic  ?  Trans- 
form so  as  to  remove  the  second  term. 


THE   ALGEBRAIC   SOLUTION    OF    EQUATIONS          231 

Ex.  5.  In  V,  Ex.  1,  let  d  =  CM,  z  =  cX,  where  /u  and  X  are  numbers  in 
the  domain  O(i),  or  in  any  other  domain.  Show  that  x5  +  ex  +  d  =  0  is 
always  metacyclic  when 

5VX 
(X-l)*(X2-6X+25)' 


(X  -  1)4(X2  -  6  X  +  25) 

Ex..  6.  Construct  the  metacyclic  quintic  in  which  /*  =  V2,  X  =  V6. 
See  Ex.  5. 

Ex.  7.   Is  x5  +  x  +  1  =  0  metacyclic  ? 

Ex.  8.  There  is  a  theorem  to  the  effect  that  all  irreducible,  metacyclic 
equations  of  the  sixth  degree  in  a  domain  ft  may  be  found  by  adjoining  to 
ft  a  square  root  and  then  forming  in  the  enlarged  domain  all  cubic  equa- 
tions. See  Weber's  Algebra,  Vol.  II,  1896,  p.  296.  Accordingly,  adjoining 
\/2  to  ft(i),  we  may  write  x3  +  x  -i-  1  +  V2  =  0  and  obtain,  by  transposing 
V2  and  squaring,  the  metacyclic  sextic  yf  +  2  x*  +  2  x9  +  x2-\-x— 1=0. 
Derive  similar  equations,  using  the  radical  \/3. 


Ex.  9.    Show  that  x5  +  5px*  +  10 p^xs  +  10^3x2  +  5p*x  +  p6  —  1  =  0  is 
metacyclic.     Also  determine  its  Galois  group. 
Increase  its  roots  by  p. 

Ex.  10.   Show  that  y5  +  pys  +  %p2y  +  r  =  0  is  metacyclic. 
Take  y  _  z  — Vf 

Ex.  11.  Prove  that  equation  V  in  Ex.  1  can  have  no  rational  root  when 
c  =  ±  1.  Then  prove  that,  ifx5  ±x  +  d  =  0  is  solvable,  it  is  reducible. 

Ex.  12.  Show  that  x5  —  A  =  0,  where  A  is  not  a  perfect  fifth  power,  is 
metacyclic  and  has  the  group  6r2o(5)  in  the  domain  0(1,  A)- 

Ex.  13.  Prove  that  an  irreducible  equation /(x)  =0  of  the  prime  degree 
n  can  become  reducible  by  adjoining  a  radical  \/«,  where  m  is  prime,  only 
when  m  =  n. 

Let  y™  —  a  =  0  I 

be  irreducible,  let  it  have  the  roots  7,  «ry,  •••,  wm-ly,  where  w  is  a  complex 
with  root  of  unity-     Let  f(x)  =  0  become  reducible  when  7  is  adjoined  to 

0,  so  that  f(f\  —  f,(f  ->A  .  f«(t  -v\  II 

J\x)—  Ji\.x'i  i)  vav*)  1)1 

the  coefficient  of  the  highest  power  of  x  in  each  polynomial  being  unity. 
We  may  consider  I  and  II  as  equations  in  the  same  domain,  having  the 


232  THEORY   OP   EQUATIONS 

root  7  in  common.  Then  II  must  be  satisfied  by  all  the  roots  of  t 
Multiplying  together  the  members  of  the  m  equations  thus  obtained,  we  get 

£«>sJU«).JK«), 

where  FI  (x)  =  /i  (x,  y)  •  fi  (x,  uy)  —fi(x,  um~li)  , 

F2(x)=f2(x,  7)  -MX,  «7)  -MX,  w-  »?). 

-Fi(x)  and  FZ(X)  are  respectively  of  the  degrees  mni  and  mw2;  their 
coefficients,  being  symmetric  functions  of  the  roots  of  I,  lie  in  ft.  Since 
/(x)  is  irreducible  and  m  and  ?z  are  both  prime,  we  must  have 


Ex.  14.  Show  that  in  Ex.  13/(x)  =/i(x,  7)  •  /i(x,  wy)  •••  /i(x,  w"-^), 
where  /i(x,  7)  is  irreducible  in  the  domain  0(w,  7),  and  is  linear  with 
respect  to  x. 

Ex.  15.    Show  that  if  /i(x,  7)  =  0  yields  in  Ex.  14 

«o  =  C0  +  Ci7  +  C272  -\  -----  h  Cn-17"-1, 

then  «!  =  c0  +  wy  +  C2w272  +  ----  h  cn_iwn-17B~1) 

etc.,  where  Oo,  «i,  etc.,  are  roots  of  /(x)  =  0,  and  c0,  Ci,  •••,  cn_i  are  num- 
bers in  0.  Show  that  the  difference  of  two  roots  of  /(x)  =  0  cannot  be  a 
number  in  ii. 

Ex.  16.  Prove  that  an  irreducible  solvable  quintic  with  real  coefficients 
cannot  have  three  real  roots  and  two  complex  roots. 

Show  that  the  Galois  group  (1)  must  be  of  the  fifth  degree  ;  (2)  can- 
not be  £i2<5>,  #6(5)I,  G6(5>H  (Ex.  5,  §  104)  ;  (3)  cannot  be  £5<5>,  §  171  ; 
(4)  to  test  Cr2o(5),  take  ?/2  in  Ex.  1,  which  admits  it.  If  any  two  roots, 
say  «0  and  «i,  are  assumed  to  be  conjugate  imaginaries,  then 

y  =  OoA  +  aiB  +  C, 

where  A,  B,  C  are  real  values.  Since  A  =  «4  —  «2  —  «3,  B  =  «2  —  03  —  «4, 
we  cannot  have  A  =  B,  because  that  would  make  «2  =  <*4-  Thus,  we  see 
that  y  cannot  be  real.  Consequently  y2  cannot  be  real,  unless  y  is  a  pure 
imaginary.  Hence  ?/=(«o—  "i)(«4  —  «-2)-  That  y'2  may  lie  in  O,  we  must 
have  y  =  iVf  •  Vg  and  «0  —  «i  =  i  V/,  #4  —  «2  =  V</,  where  /  and  gr  are 
positive  numbers  in  0.  But  by  Ex.  15,  /and  g  cannot  be  perfect  squares. 
By  Exs.  13,  14,  15  we  see  that  the  roots  of  the  given  quintic  are  numbers 
in  the  domain  12(Wi  ),  where  w  is  a  complex  fifth  root  of  unity  and  7  is  a 
root  of  the  irreducible  equation  y&  —  a  —  0.  Hence  V/  and  Vg  do  not 
lie  in  fl((0i  y)  and  the  equations  «0  —  «i  =  iVf,  «4  —  a2  =  Vg  are  impossible. 
Consequently  (?2o(5)  is  not  the  group,  §  155,  B. 


THE    ALGEBRAIC    SOLUTION   OF    EQUATIONS          233 

(5)  Since  Gw^  does  not  alter  y2,  it  is  not  the  group. 

(6)  Hence  the  group  must  be  £120<5>  or  (?6o(5),  both  insolvable. 

For  different  proofs  see  Weber's  Algebra,  Vol.  I,  p.  669,  and  Weber's 
JSncyklopadie  der  Elementaren  Algebra  und  Analysis,  p.  327. 

Ex.  17.  Show  that  x5  —  4  a;  —  2  =  0  has  two  complex  roots  and  is 
insolvable.  For  the  approximate  values  of  the  real  roots,  see  §  26. 

Ex.  18.    Show  that  x5  -  16  x2  +  2  x  +  6  =  0  is  insolvable. 
Ex.  19.    Show  that  x5  +  1  +  i  =  0  is  metacyclic. 

Ex.  20.    Determine  which  of  the  following  are  metacyclic : 
(a)  a5  +  5  x  +  3  i  =  0.  &-\  _  Q 

(6)  s#  -  2  fz  +  7  =  0.  x  ~  l 

(d)  x*  -  27  x*  +  3  x  +  6  =  0. 

201.  Historical  References.  For  the  development  of  the  earlier  and 
more  elementary  parts  of  the  theory  of  equations  consult  the  histories  of 
mathematics  written  by  Ball,  Fink,  Marie,  Zeuthen,  and  Cajori,  and  the 
"  Notes  "  at  the  close  of  the  first  volume  of  Burnside  and  Panton's  Theory 
of  Equations.  Or,  better  yet,  consult  the  monumental  work  by  Moritz 
Cantor,  entitled  Vorlesungen  uber  Geschichte  der  Mathematik.  For  the 
later  developments,  read  C.  A.  Bjerknes'  Niels-Henrik  Abel  (Paris,  1885); 
Evariste  Galois'  CEuvres,  edited  by  Picard  (1897) ;  H.  Burkhardt's 
"Anfange  der  Gruppentheorie  und  Paolo  Ruffini,"  in  the  Zeitsch.  fur 
Mathematik  und  Physik  (Vol.  37,  Sup.,  pp.  119-159,  1892).  Read  articles 
in  the  Bulletin  of  the  American  Mathematical  Society,  by  James  Pierpont, 
on  Lagrange's  place  in  the  theory  of  substitutions  (Vol.  1,  pp.  2,  196-204, 
1895),  on  the  early  history  of  Galois'  theory  of  equations  (Vol.4,  pp.  332- 
337,  1898),  on  Galois'  Collected  Works  (Vol.  5,  pp.  296-300,  1899)  ;  by 
G.  A.  Miller,  a  report  on  recent  progress  in  the  theory  of  the  groups  of  a 
finite  order  (Vol.  5,  pp.  227-249,  1899)  ;  by  Henry  B.  Fine,  on  "Kronecker 
and  his  Arithmetical  Theory  of  the  Algebraic  Equation  "  (Vol.  1,  pp.  173- 
184,  1892).  Consult  also  James  Pierpont,  "  Zur  Geschichte  der  Gleichung 
des  V.  Grades  (bis  1858),"  in  Monatshefte  fur  Mathematik  und  Physik 
(Vol.  6,  pp.  15-68,  1895)  ;  G.  A.  Miller  on  the  history  of  several  funda- 
mental theorems  in  the  theory  of  groups  of  a  finite  order,  in  the  American 
Mathematical  Monthly  (Vol.  8,  pp.  213-216,  1901)  ;  Felix  Klein,  Vorle- 
sungen uber  das  Ikosaeder  (1884),  also  Lectures  on  Mathematics  (the 
Evanston  Colloquium,  1894)  ;  B.  S.  Easton,  The  Constructive  Develop- 
ment of  Group-theory  (Philadelphia,  1902). 


ANSWERS 


§  6,  Ex.  4:  47112. 
Ex.  5  :  -  252493. 
Ex.  6:  J(l  ±v^2 
§  14,  Ex.  3 :    -  i,  ±  V5. 
Ex.4:    -|,  - 
Ex.  5  :   _  3,  _  5,  _  5. 

Ex.  6:   |,  *,  -3,  -3. 
§  15,  Ex.  2  :  a2  -  2  6. 

Ex.  6 :  c  —  ab. 

Ex.  7  :  a2  -  2  6. 

Ex.  8 :  3  c  —  ab. 

Ex.  9 :  62  -  2  ac  +  2  d. 

Ex.  12  :    -  £,  rfo. 
§  21,  Ex.  4 :   -  1  triple,  f  double. 
§  29,  Ex.  7 :      z*  -  60  ^  -f  700  x  - 
100  =  0. 

Ex.  8 :       sc4  -  3  x3  +  768  a;  + 

1024  =  0. 
§  31,  Ex.  3 :  am  =  0.  

Ex.  6=c,±l,-&±V62-4a2. 

§  34,  Ex.  1 :  If  =  |,  £  =  -  12. 

Ex.  2  :  //  =  -  Jp,  £  =  -  135, 

§  35,  Ex.  5  :      z3  -  42  z2  +  441  z  + 

1388  =  0.         I"^^$T 
§  37,  Ex.  2 :  Two  of  the  roots  of  I 
are  equal,  or  the  three  are 
in  arithmetical  progression. 
Ex.  3 :   -  2376. 
Ex.  4 :  0. 


39,  Ex.  2 :  (1)  41,  Iff 

(2)  4J,  1|J. 

(3)  37,  7. 

(4)  Si,  3i. 

41,  Ex.3:   (1)   ^and  -5. 

(2)  4  and  -  f 

(3)  12  and  -  $f . 

42,  Ex.  3:  (1)  Between  8  and  9, 

—  1  and  —  2,  -  4  and  —  6. 

(2)  Between  3  and  4, 

4  and  5,  —  4  and 
—  5,  —5  and  —6. 

(3)  Between  1  and  2, 

2  and  3,  5  and  6, 

6  and  7. 
156,  31. 
(1)  two  real  2. +,  - 


44,  Ex.  2 : 
49,  Ex.  5 : 

2.+ 


(2)  3.21,  3.22,  -  17:4. 

(3)  three  real. 

§  50,  Ex.  1:  H=Q,    Q_=  25,    /  = 

289,  «7=-940,  D  =  +  . 
§  56,  Ex.1:   (1)  1.35759- 

(2)  -1.53172- 

(3)  .885119- 

-  1.46057- 

(4)  1.3518- 

(5)  1.51851-. 

-  .50849- 

-  1.24359- 
§67,  Ex.  1:   -1,  -i 


236 


ANSWERS 


§71,  Ex.1: 

(3  c  -  06)  (a2  -  2  &) 
(62-2ac)(a26-&2-3«o)' 

Ex.  5  :  For/(x)  =  0,  ar'a2  - 
2  «22  —  «i#3  +  4  a4. 

Ex.  6 :   —  0203  +  3  aia4—  5  as. 
Yes. 

Ex.  7 :   ^  (&!2  - 

Ex.  8 :  24.  -  \  *} 
Ex.   11 :    x3"-  «2x2  + 

4  «4)  x  —  as2—  ai2a4  +  4  a«a4 
=  0. 
Ex.    15 :     -  2  ai8  +  9  aia2  + 

27  as. 

Ex.17:   x8.~.12/  +  VZ>  =  0. 
§  77,  Ex.  3 :  The  roots  of  49  a2  - 

163  a  +  283  =  0. 
Ex.  5:  n. 
Ex.  6 :  0. 

§  93,  Ex.  1:  (123) (412)  (256). 

§  113,  Ex.  4  :   £4(4)  II. 

Ex.  5  :   Q4W  II. 

Ex.  6:   #4(4)II. 

§  123,  Ex.  2 :   (c)  \/5. 


(e)  V= 
Ex.  3  : 
§128,  Ex.2:    (1),    (3),    (4),   (7), 

(8),  are  reducible. 

§  133,  Ex.  7 :  0  =  (<  +  V2  +  V3). 

§  135,  Ex.  7:  Try  N=  a  +  a*+  ct? 

=  a4  +  («*)« 


141,  Ex.  2  :  Let  x  =  4  BI  +  1. 

142,  Exs.  2,  3  :  No. 


§  148,  Ex.  3 
§  159,  Ex.  8 


§  163,  Ex.  2 


§  188,  Ex.  1 
§  199,  Ex.  1 


(««3). 
(a)  P=l. 
(6)  P=  £2<2>. 

(C)   P  =  04(0  II. 
(d)P=l. 

(e)  P  =  £3<3>. 
(SOP=£4«>I. 
(0  P=  04  W  III. 
(&)  P  =  the  product 


each    group 
involving  distinct 
roots  of  its  own  as 
elements. 
(Z)  P  =  #4(4)  HI,  or 

a  sub-group. 
(m)  £6<3). 
(w)  Let  x  =  y  —  1. 
:    4  «!  =  1  -  z  -  ID  + 


w\  —  z\. 
4:  a  —  1  +  z  +  w  — 

Wi  +  Zi. 


INDEX 


(The  numbers  refer  to  pages.) 


Abel,  233. 

Abelian  equations,  210. 

Abelian  groups,  210. 

Adjunction,  135,  151,  219. 

Algebraic  numbers,  136. 

Algebraic    solutions,    60,    219 ;    of 

cubic  and  quartic,  68. 
Angle,  trisection  of,  207,  208. 

Bachmann,  206. 

Ball,  233. 

Beman,  W.  W.,  206. 

Binomial  equations,  74,  219. 

Biquadratic,  see  Quartic. 

Bjerknes,  C.  A.,  233. 

Budan,  50. 

Burkhardt,  233. 

Burnside  and  Panton,  79,  233. 

Cantor,  M.,  233. 
Cardan's  formula,  69. 
Carvallo,  M.  E.,  67. 
Cole,  F.  N.,  189. 

Complex  roots,  6,  42,  58,  67,  232. 
Composite  sub-groups,  123. 
Conjugate  sub-groups,  122. 
Constructions   by  ruler   and   com- 
passes, 202. 
Continuity  of /(x),  25. 
Cross-ratio,  100,  127. 
Cube,  duplication  of,  207. 


Cubic,  algebraic  solution,  68  ;  cyclic, 
196 ;  equation  of  squared  differ- 
ences of,  38  ;  irreducible  case,  69, 
208 ;  nature  of  roots,  41 ;  reduc- 
ing cubic,  72  ;  removal  of  second 
term,  36. 

Cyclic  equations,  187,  196,  198, 
220. 

Cyclic  function,  115,  127,  128,  133. 

Cyclic  group,  115,  128,  132,  133. 

Cyclotomic  equations,  142,  198. 

Delian  problem,  207. 

De  Moivre's  theorem,  24. 

Descartes,  50. 

Descartes'  Rule  of  Signs,  7,  60. 

Dialytic  method  of  elimination,  95. 

Discriminant,    110 ;    of    quadratic, 

97  ;  of  cubic,  40 ;  of  quartic,  59 ; 

of  /(»)  =  0,  96,  97. 
Division  of  the  circle,  75,  203-206. 
Domain,   defined,    134 ;    conjugate, 

143 ;  degree  of,  142  ;  Galois,  153  ; 

normal,  142,  150  ;  primitive,  144  ; 

substitutions  of,  160. 
Duplication  of  the  cube,  207. 

Easton,  B.  S.,  233. 
Eisenstein's  theorem,  141. 
Eliminants,  92. 
Equal  roots,  21,  63,  142. 


237 


238 


INDEX 


Equations,  Abelian,  210  ;  algebraic, 
2 ;  algebraic  solution  of,  219 ; 
binomial,  74,  219 ;  cubic,  36,  38, 
41,  68,  69,  72,  196,  208;  cyclic, 
187,  220;  cyclotomic,  142;  irre- 
ducible, 137  ;  metacyclic,  223 ; 
quadratic,  184 ;  quartic,  185 ; 
quintic,  186,  227,  229,  232,  233; 
reciprocal,  33. 

Euler's  cubic,  71. 

Kuler's  method  of  elimination,  94. 

Euler's  solution  of  quartic,  71. 

Fine,  H.  B.,  233. 

Fink,  233. 

Fourier,  50. 

Function,  def.  1 ;  alternating,  115 ; 
"belongs  to,"  115,  124,  125; 
cyclic,  115,  127, 128  ;  derived,  18  ; 
Sturm's,  50 ;  resolvents  of  La- 
grange,  129 ;  symmetric,  13,  84, 
114. 

Galois,  143,  176,  233. 

Galois'  theory  of  numbers,  134  ;  do- 
main, 153 ;  resolvent,  155,  156, 
reduction  of,  174,  178 ;  groups, 
164,  determination  of,  169. 

Gauss,  26,  206 ;  Lemma,  138. 

Graphic  representation,  15,  28,  75. 

Groups,  112  ;  Abelian,  210  ;  alter- 
nating, 115 ;  composite,  123 ; 
cyclic,  115,  128,  132,  133 ;  degree 
and  order  of,  113  ;  Galois,  164  ; 
index  of,  122;  list  of,  118,  119; 
normal  sub-groups,  122 ;  124 ; 
primitive  and  imprimitive,  116 ; 
simple,  122 ;  sub-groups,  120 ;  sym- 
metric, 1 14 ;  transitive  and  in- 
transitive, 116. 

Hermite,  137. 

Historical  references,  233. 

Homograph ic  transformation,  99. 


Homer's  method,  63. 

Imaginary    roots,    6,    42,    58,    67, 

232. 

Imprimitive  group,  116,  127. 
Invariant  sub-groups,  122. 
Irreducible  case  in  cubic,  69,  208. 

Klein,  F.,  206,  233. 
Kronecker,  187,  233. 

Lagrange,  233 ;  resolvents  of,  129 ; 

theorem  of,  176. 
Lindemann,  137. 

Marie,  233. 

Matthiessen,  L.,  186. 

McClintock,  E.,  67,  230. 

Metacyclic  equations,  223,  227. 

Miller,  G.  A.,  233. 

Moritz,  26. 

Multiple  roots,  21,  53,  142.    ^ 

Netto,  189,  218,  228,  229. 

Newton,  50. 

Newton's  formula  for  sums  of  pow- 
ers, 84. 

Newton's  method  of  approximation, 
66. 

Normal  domain,  142,  145,  150. 

Normal  equations.  149,  151. 

Normal  sub-groups,  122 ;  of  prime 
index,  124. 

Numbers,  algebraic,  136  ;  conjugate, 
144 ;  primitive,  144,  147 ;  tran- 
scendental, 137. 

Pan  ton,  see  Burnside  and  Pan  ton. 

Picard,  233. 

Pierpont,  J.,  233. 

Primitive  congruence  roots,  199. 

Primitive  domains,  144,  147. 

Quadratic  equation,  184. 


INDEX 


239 


Quartic,  cyclic,  198 ;  Euler's  solu- 
tion, 71  ;  groups  of,  172,  173 ;  in 
the  Galois  theory,  185  ;  nature  of 
roots,  56 ;  removal  of  second 
term,  37 ;  symmetric  functions 
of  roots,  91 ;  when  solvable  by 
square  roots,  72. 

Quintic,  186,  227,  229,  232,  233. 

Radicals,  solution  by,  60. 

Reciprocal  equations,  33 ;  depres- 
sion of,  81. 

Reducibility,  134,  135,  139. 

Reducing  cubic,  72. 

Regular  polygons,  inscription  of, 
20. 

Resolvents  of  Lagrange,  129. 

Resultants,  92. 

Rolle's  theorem,  49. 

Roots,  2 ;  complex,  6,  42,  58,  67, 
232  ;  fractional,  61 ;  fundamental 
theorem,  26 ;  incommensurable, 
61 ;  integral,  62  ;  multiple  or  equal 
roots,  21,  53,  142 ;  of  unity,  76, 
198  ;  primitive,  78  ;  primitive  con- 
gruence roots,  199  ;  reciprocal,  33. 

Ruffini,  P.,  233. 

Runge,  C.,  229. 

Self-conjugate  sub-groups,  122. 
Simple  groups,  122. 
Smith,  D.  E.,  206. 
Solvable  equations,  223. 
Sturm,  50. 


Sturm's  theorem,  50,  51  ;  applied  to 
quartic,  56. 

Sub-groups,  120  ;  index  of,  122  ;  of 
prime  index,  124. 

Substitutions,  104 ;  cyclic,  107 ; 
even  and  odd,  111;  identical, 
106 ;  inverse,  106 ;  laws  of,  105 ; 
product  of,  105. 

Substitution  groups,  see  Groups. 

Sylvester,  50. 

Sylvester's  method  of  elimination, 95. 

Symmetric  functions,  13,  84,  114 ; 
fundamental  theorem,  87  ;  elimi- 
nation by,  93. 

Symmetric  group,  114. 

Synthetic  division,  3. 

Taylor's  theorem,  19. 

Transcendental  numbers,  137. 

Transpositions,  109. 

Trigonometric  solution  of  irreduci- 
ble case,  70 ;  of  binomial  equa- 
tions, 74,  82,  83. 

Trisecting  an  angle,  207,  208. 

Tschirnhausen's  transformation,  99, 
102. 

Unity,  roots  of,  76,  198 ;  primitive 
roots  of,  78. 

Waring,  50. 

Weber,  H.,  29,  134,  228,  231. 

Zeuthen,  233. 


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